Answers and detailed solutions 45-minute test (1 period) – Question 4 – Chapter IV – Calculus 12.
Topic
Question 1. Given two complex numbers ({z_1} = 9 – i,,,,{z_2} = – 3 + 2i). Calculate the value of (left| {dfrac{{{z_1}}}{{{z_2}}}} right|) equal to /
A. (dfrac{{2sqrt {154} }}{{13}}). B. (dfrac{{616}}{{169}}).
C. (dfrac{{82}}{{13}}). D. (sqrt {dfrac{{82}}{{13}}} ).
Verse 2. Given two complex numbers ({z_1} = a + bi,,,{z_2} = c + di)z. Find the real part of the complex number ({z_1}. {z_2}).
A. The real part of the complex number ({z_1}.{z_2}) is ac + bd.
B. The real part of the complex number ({z_1}. {z_2}) is ac – bd .
C. The real part of the complex number ({z_1}.{z_2}) is ad + bc.
D. The real part of the complex number ({z_1}.{z_2}) is ad – bc
Verse 3. Given a complex number (z = – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i). Then the complex number ({left( {overline z } right)^2}) is equal to ;
A. ( – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i).
B. (sqrt 3 – i).
C. ( – dfrac{1}{2} – dfrac{{sqrt 3 }}{2}i).
D. (1 + sqrt 3 i).
Verse 4.Assume that A, B are respectively the representation of complex numbers ({z_1} = {a_1} + {b_1}i,,,,{z_2} = {a_2} + {b_2}i ). Then the length of the vector (overrightarrow {AB} ) is equal to ;
A. (|{z_1} + {z_2}|).
B. (|{z_1}| + |{z_2}|).
C. (|{z_1}| – |{z_2}|).
D. (|{z_1} – {z_2}|).
Question 5. The modulus of the complex number z satisfying (dfrac{{2 + i}}{{1 – i}}z = dfrac{{ – 1 + 3i}}{{2 + i}}) is:
A. (sqrt 5 ) B. (dfrac{{sqrt 5 }}{5})
C. (dfrac{{2sqrt 5 }}{5}) D. (dfrac{{3sqrt 5 }}{5}).
Verse 6. Calculate the following complex number: (z = {left( {1 + i} right)^{15}}).
A. (z = – 128 + 128i).
B. (z = 128 – 128i).
C. (z = 128 + 128i).
D. (z = – 128 – 128i).
Verse 7. Let the complex number z = a + bi. Then the number (dfrac{1}{2}left( {z + overline z } right)) is:
A. A pure imaginary number.
B. 2a.
C. i.
D. a.
Verse 8. For complex numbers ({z_1} = 2 – 5i,,,,{z_2} = – 2 – 3i). Let’s calculate (|{z_1} – {z_2}|).
A. (2sqrt 5 ) B. 20
C. 12 D. (2sqrt 3 ).
Verse 9. Given a complex number z satisfying (left( {3 – 2i} right)z = 4 + 2i). Find the conjugate complex number of z.
A. (overline z = 4 – 2i).
B. (overline z = dfrac{8}{{13}} + dfrac{{14}}{{13}}i).
C. (overline z = 3 + 2i).
D. (overline z = dfrac{8}{{13}} – dfrac{{14}}{{13}}i).
Question 10. Solving the equation ({z^2} – 6z + 11 = 0), we have a solution of :
A. (z = 3 + sqrt 2 i).
B. (z = 3 – sqrt 2 i).
C. (left[begin{array}{l}z=3+sqrt2iz=3–sqrt2iend{array}right)[begin{array}{l}z=3+sqrt2iz=3–sqrt2iend{array}right)
D. Another result.
Verse 11. Given two complex numbers (z = a + bi,,,,z’ = a’ + b’i). Choose the correct formula.
A. (z + z’ = left( {a + b} right) + left( {a’ + b’} right)i).
B. (z – z’ = left( {a + a’} right) – left( {b + b’} right)i).
C. (z.z’ = left( {aa’ – bb’} right) + left( {ab’ + a’b} right)i).
D. (z.z’ = left( {aa’ + bb’} right) – left( {ab’ + a’b} right)i).
Verse 12. Let z = 1 + 2i. The real and imaginary parts of the complex number (w = 2z + overline z ) are:
A. 3 and 2.
B. 3 and 2i.
C. 1 and 6.
D. 1 and 6i.
Verse 13. The solution of the system of equations (left{ begin{array}{l}x + 2y = 1 + i3x + iy = 2 – 3iend{array} right.) is:
A. (left{ begin{array}{l}x = 1 + iy = iend{array} right.).
B. (left{ begin{array}{l}x = iy = 1 + iend{array} right.).
C. (left{ begin{array}{l}x = 1 – iy = iend{array} right.).
D. (left{ begin{array}{l}x = iy = 1 – iend{array} right.).
Verse 14. Find a complex number with a real part of 12 and a modulus of 13.
A. (5 pm 12i).
B. 12 + 5i.
C. (12 pm 5i).
D. (12 pm i).
Verse 15. The equation ({z^2} – 2z + 3 = 0) has the following solutions:
A. (2 pm 2sqrt 2 i).
B. ( – 2 pm 2sqrt 2 i).
C. ( – 1 pm 2sqrt 2 i).
D. (1 pm sqrt 2 i).
Verse 16. The set of points representing complex numbers z satisfying (|overline z + 3 – 2i| = 4) is:
A. A circle with center I(3 ; 2) has radius R = 4.
B. The circle with center I(3 ; -2) has radius R= 4.
C. The circle with center I(-3 ; 2) has radius R = 4.
D. The circle with center I(- 3; -2) has radius R = 4.
Verse 17. Two points representing two conjugate complex numbers (z = 2 + 2i,,,overline z = 2 – 2i) are symmetric through :
A. Vertical axis.
B. Horizontal axis.
C. Origin coordinates.
D. Grade A(2; -2).
Verse 18. Given a complex number (z = rleft( {cos dfrac{pi }{2} + isin dfrac{pi }{2}} right)). Choose 1 accumen of z:
A. ( – dfrac{pi }{2}) B. ( – dfrac{{3pi }}{2})
C. (dfrac{{3pi }}{2}) D. (pi ).
Verse 19. Modulus of the sum of two complex numbers ({z_1} = 3 – 4i,,,,{z_2} = 4 + 3i):
A. (5sqrt 2 ) B. 10
C. 8 D. 50.
Verse 20. Given a complex number (z = – rleft( {cos varphi + isin varphi } right)). Find an accumen of z ?
A. ( – varphi ).
B. (varphi + 2pi ).
C. (varphi – 2pi ).
D. (varphi + pi ).
Verse 21. Calculate (z = dfrac{{5 + 5i}}{{3 – 4i}} + dfrac{{20}}{{4 + 3i}}).
A. z = 3 – i.
B. z = 3 + i.
C. z = – 3 – i.
D. z = – 3 + i.
Question 22.The set of points representing complex numbers z satisfying (|z + 1 + i|, le 2) is;
A. A circle with center I(1 ; 1) and radius R = 2.
B. A circle with center I(1; 1) and radius R = 2.
C. Circle with center I(- 1 ; – 1) radius R = 2.
D. Circle with center I(- 1 ; – 1) radius R = 2.
Verse 23. The trigonometric form of the complex number z = i – 1 is:
A. (z = sqrt 2 left( {cos dfrac{{3pi }}{4} – isin dfrac{{3pi }}{4}} right)).
B. (z = 2left( {cos dfrac{{3pi }}{4} + isin dfrac{{3pi }}{4}} right)).
C. (z = sqrt 2 left( {cos dfrac{{ – pi }}{4} + isin dfrac{{ – pi }}{4}} right)).
D. (z = sqrt 2 left( {cos dfrac{{3pi }}{4} + isin dfrac{{3pi }}{4}} right)).
Verse 24. In the complex plane, the points A, B are the representation points of ({z_1} = 2 – 4i,,,,{z_2} = 4 + 5i, respectively). The midpoint of AB has coordinates:
A. (Aleft( {3;dfrac{3}{2}} right)).
B. (Aleft( {3;1} right)).
C. (Aleft( {3;dfrac{1}{2}} right)).
D. (Aleft( {6;1} right)).
Verse 25. Given a complex number z such that (left( {3 + 2i} right)z + {left( {2 – i} right)^2} = 4 + i). The modulus of the complex number (w = left( {z + 1} right)overline z ) is:
A. 2 B. 4
C. 10 D. (sqrt {10} ).
Detailed explanation
first | 2 | 3 | 4 | 5 |
EASY | REMOVE | A | EASY | OLD |
6 | 7 | 8 | 9 | ten |
REMOVE | EASY | A | EASY | OLD |
11 | twelfth | 13 | 14 | 15 |
OLD | A | OLD | OLD | EASY |
16 | 17 | 18 | 19 | 20 |
A | REMOVE | REMOVE | A | EASY |
21 | 22 | 23 | 24 | 25 |
A | EASY | EASY | OLD | EASY |
Detailed explanation
Question 1: EASY
(begin{array}{l}{z_1} = 9 – i;{z_2} = – 3 + 2idfrac{{{z_1}}}{{{z_2}}} = dfrac{{ left( {9 – i} right)left( { – 3 – 2i} right)}}{{9 – 4{i^2}}},,,,,, , = dfrac{{ – 27 + 2{i^2} – 15i}}{{13}} = – dfrac{{29}}{{13}} – dfrac{{15}}{{13 }}i Rightarrow left| {dfrac{{{z_1}}}{{{z_2}}}} right| = sqrt {{{left( { – dfrac{{29}}{ {13}}} right)}^2} + {{left( { – dfrac{{15}}{{13}}} right)}^2}};;; ;;;;;;;;;, = sqrt {dfrac{{82}}{{13}}} end{array})
Verse 2: REMOVE
Question 3: A
(begin{array}{l}z = – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i Rightarrow {left( {overline z } right )^2} = {left( { – dfrac{1}{2} – dfrac{{sqrt 3 }}{2}i} right)^2},,,, ,,,,,,,,,,,,, = dfrac{{{{left( {1 + sqrt 3 i} right)}^2 }}}{4},,,,,,,,,,,,,,,,, = dfrac{{ – 2 + 2sqrt 3 i}}{4},,,,,,,,,,,,,,,,, = – dfrac {1}{2} + dfrac{{sqrt 3 }}{2}iend{array})
Question 4: EASY
Verse 5 C
(begin{array}{l}dfrac{{2 + i}}{{1 – i}}z = dfrac{{ – 1 + 3i}}{{2 + i}} Leftrightarrow z = dfrac{{left( { – 1 + 3i} right)left( {1 – i} right)}}{{{{(2 + i)}^2}}} Leftrightarrow z = dfrac{{2 + 4i}}{{3 + 4i}} Leftrightarrow z = dfrac{{left( {2 + 4i} right)left( {3 – 4i} right)} }{{9 – 16{i^2}}} Leftrightarrow z = dfrac{{6 – 16{i^2} + 4i}}{{25}} Leftrightarrow z = dfrac{{ 22}}{{25}} + dfrac{4}{{25}}i Rightarrow left| z right| = dfrac{{2sqrt 5 }}{5}end{array} )
Sentence 6: REMOVE
(begin{array}{l}z = {(1 + i)^{15}} = {left( {1 + i} right)^{14}}(1 + i), ,,, = {({(1 + + i)^2})^7}left( {1 + i} right) = {2^7}{i^7}left( {1 + i} right),,,, = – {2^7}ileft( {1 + i} right) = 128 – 128iend{array})
Question 7: EASY
Question 8: A
({z_1} – {z_2} = left( {2 – 5i} right) – ( – 2 – 3i))(, = 4 – 2i)
( Rightarrow left| {{z_1} – {z_2}} right| = 2sqrt 5)
Question 9: EASY
(begin{array}{l}left( {3 – 2i} right)z = 4 + 2i Leftrightarrow z = dfrac{{4 + 2i}}{{3 – 2i}} Leftrightarrow z = dfrac{{(4 + 2i)(3 + 2i)}}{{9 – 4{i^2}}},,,,,,,, ,,,, = dfrac{{12 + 4{i^2} + 14i}}{{13}},,,,,,,,, ,,, = dfrac{8}{{13}} + dfrac{{14}}{{13}}i Rightarrow overline z = dfrac{8}{{13}} – dfrac{{14}}{{13}}iend{array})
Question 10:
(begin{array}{l}{z^2} – 6z + 11 = 0 Leftrightarrow left( {{z^2} – 6z + 9} right) + 2 = 0 Leftrightarrow {(z – 3)^2} + 2 = 0 Rightarrow left[begin{array}{l}z–3=isqrt2z–3=–isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=3+isqrt2z=3–isqrt2end{array}rightend{array})[begin{array}{l}z–3=isqrt2z–3= –isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=3+isqrt2z=3–isqrt2end{array}rightend{array})
Question 11:
Question 12: A
({rm{w}} = 2z + overline z = 2(1 + 2i) + (1 – 2i) )(,= 3 + 2i)
real part: 3 , imaginary part: 2
Verse 13:
(left{ begin{array}{l}x + 2y = 1 + i3x + iy = 2 – 3iend{array} right. )
(Leftrightarrow left{ begin{array}{l}x = 1 + i – 2y{rm{ (1)}}3x + iy = 2 – 3i{rm{ (2)}} end{array} right.)
Substituting (1) into (2) we get:
(begin{array}{l}3(1 + i – 2y) + iy = 2 – 3i Leftrightarrow ( – 6 + i)y = – 1 – 6i Leftrightarrow y = dfrac{{ – 1 – 6i}}{{ – 6 + i}} Leftrightarrow y = dfrac{{left( { – 1 – 6i} right)left( { – 6 – i} right)}} {{36 – {i^2}}} = iend{array})
Replace y = i in (1) ( Rightarrow x = 1 – i)
Verse 14:
With the real part equal to 12, the complex number z has the form (z = 12 + bi)
(begin{array}{l}left| z right| = 13 Rightarrow left| {12 + bi} right| = 13 Leftrightarrow sqrt {{{12}^2} + { b^2}} = 13 Leftrightarrow {b^2} = 25 Leftrightarrow left[begin{array}{l}b=5Rightarrowz=12+5ib=–5Rightarrowz=12–5iend{array}rightend{array})[begin{array}{l}b=5Rightarrowz=12+5ib= –5Rightarrowz=12–5iend{array}rightend{array})
Question 15: EASY
(begin{array}{l}{z^2} – 2z + 3 = 0 Leftrightarrow left( {{z^2} – 2z + 1} right) + 2 = 0 Leftrightarrow {left( {z – 1} right)^2} + 2 = 0 Leftrightarrow {left( {z – 1} right)^2} = – 2 Rightarrow left[begin{array}{l}z–1=isqrt2z–1=–isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=1+isqrt2z=1–isqrt2end{array}rightend{array})[begin{array}{l}z–1=isqrt2z–1= –isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=1+isqrt2z=1–isqrt2end{array}rightend{array})
Verse 16: A
Verse 17: REMOVE
Verse 18: REMOVE
Question 19: A
(begin{array}{l}{z_1} + {z_2} = 3 – 4i + 4 + 3i = 7 – i Rightarrow left| {{z_1} + {z_2}} right| = 5 sqrt 2 end{array})
Question 20: EASY
Question 21: A
(begin{array}{l}z = dfrac{{5 + 5i}}{{3 – 4i}} + dfrac{{20}}{{4 + 3i}},, ,, = dfrac{{5left( {1 + i} right)left( {3 + 4i} right)}}{{9 – 16{i^2}}} + dfrac{{ 20left( {4 – 3i} right)}}{{16 – 9{i^2}}},,,, = dfrac{{5(3 + 4{i^2 } + 7i) + 20(4 – 3i)}}{{25}},,,, = dfrac{{5( – 1 + 7i) + 20left( {4 – 3i} right)}}{{25}} = 3 – iend{array})
Question 22: EASY
Set (z= x+yi)
(begin{array}{l}left| {z + 1 + i} right| le 2 Rightarrow left| {x + yi + 1 + i} right| le 2 Leftrightarrow left| {left( {x + 1} right) + left( {y + 1} right)} right| le 2 Leftrightarrow sqrt {{{left( {x) + 1} right)}^2} + {{left( {y + 1} right)}^2}} le 2end{array})
So the set of points representing complex numbers z is a circle with center I(-1, -1), radius equal to 2
Question 23: EASY
Verse 24:
Question 25: EASY
(begin{array}{l}left( {3 + 2i} right)z + {left( {2 – i} right)^2} = 4 + i Leftrightarrow left( { 3 + 2i} right)z + (3 – 4i) = 4 + i Leftrightarrow left( {3 + 2i} right)z = 1 + 5i Leftrightarrow z = dfrac{{1 + 5i}}{{3 + 2i}} Leftrightarrow z = dfrac{{left( {1 + 5i} right)left( {3 – 2i} right)}}{{9 – 4{ i^2}}} Leftrightarrow z = dfrac{{13 + 13i}}{{13}} = 1 + i{rm{w}} = (z + 1)overline z = ( 2 + i)(1 – i),,,,,, = 2 – {i^2} – i = 3 – i Rightarrow left| {rm{w} } right| = sqrt {10} end{array})
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Answers and detailed solutions 45-minute test (1 period) - Question 4 - Chapter IV - Calculus 12.
Topic
Question 1. Given two complex numbers ({z_1} = 9 – i,,,,{z_2} = – 3 + 2i). Calculate the value of (left| {dfrac{{{z_1}}}{{{z_2}}}} right|) equal to /
A. (dfrac{{2sqrt {154} }}{{13}}). B. (dfrac{{616}}{{169}}).
C. (dfrac{{82}}{{13}}). D. (sqrt {dfrac{{82}}{{13}}} ).
Verse 2. Given two complex numbers ({z_1} = a + bi,,,{z_2} = c + di)z. Find the real part of the complex number ({z_1}. {z_2}).
A. The real part of the complex number ({z_1}.{z_2}) is ac + bd.
B. The real part of the complex number ({z_1}. {z_2}) is ac – bd .
C. The real part of the complex number ({z_1}.{z_2}) is ad + bc.
D. The real part of the complex number ({z_1}.{z_2}) is ad – bc
Verse 3. Given a complex number (z = – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i). Then the complex number ({left( {overline z } right)^2}) is equal to ;
A. ( – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i).
B. (sqrt 3 – i).
C. ( – dfrac{1}{2} – dfrac{{sqrt 3 }}{2}i).
D. (1 + sqrt 3 i).
Verse 4.Assume that A, B are respectively the representation of complex numbers ({z_1} = {a_1} + {b_1}i,,,,{z_2} = {a_2} + {b_2}i ). Then the length of the vector (overrightarrow {AB} ) is equal to ;
A. (|{z_1} + {z_2}|).
B. (|{z_1}| + |{z_2}|).
C. (|{z_1}| – |{z_2}|).
D. (|{z_1} – {z_2}|).
Question 5. The modulus of the complex number z satisfying (dfrac{{2 + i}}{{1 – i}}z = dfrac{{ – 1 + 3i}}{{2 + i}}) is:
A. (sqrt 5 ) B. (dfrac{{sqrt 5 }}{5})
C. (dfrac{{2sqrt 5 }}{5}) D. (dfrac{{3sqrt 5 }}{5}).
Verse 6. Calculate the following complex number: (z = {left( {1 + i} right)^{15}}).
A. (z = – 128 + 128i).
B. (z = 128 – 128i).
C. (z = 128 + 128i).
D. (z = – 128 – 128i).
Verse 7. Let the complex number z = a + bi. Then the number (dfrac{1}{2}left( {z + overline z } right)) is:
A. A pure imaginary number.
B. 2a.
C. i.
D. a.
Verse 8. For complex numbers ({z_1} = 2 – 5i,,,,{z_2} = – 2 – 3i). Let's calculate (|{z_1} – {z_2}|).
A. (2sqrt 5 ) B. 20
C. 12 D. (2sqrt 3 ).
Verse 9. Given a complex number z satisfying (left( {3 – 2i} right)z = 4 + 2i). Find the conjugate complex number of z.
A. (overline z = 4 – 2i).
B. (overline z = dfrac{8}{{13}} + dfrac{{14}}{{13}}i).
C. (overline z = 3 + 2i).
D. (overline z = dfrac{8}{{13}} – dfrac{{14}}{{13}}i).
Question 10. Solving the equation ({z^2} – 6z + 11 = 0), we have a solution of :
A. (z = 3 + sqrt 2 i).
B. (z = 3 – sqrt 2 i).
C. (left[begin{array}{l}z=3+sqrt2iz=3–sqrt2iend{array}right)[begin{array}{l}z=3+sqrt2iz=3–sqrt2iend{array}right)
D. Another result.
Verse 11. Given two complex numbers (z = a + bi,,,,z' = a' + b'i). Choose the correct formula.
A. (z + z' = left( {a + b} right) + left( {a' + b'} right)i).
B. (z – z' = left( {a + a'} right) – left( {b + b'} right)i).
C. (z.z' = left( {aa' – bb'} right) + left( {ab' + a'b} right)i).
D. (z.z' = left( {aa' + bb'} right) – left( {ab' + a'b} right)i).
Verse 12. Let z = 1 + 2i. The real and imaginary parts of the complex number (w = 2z + overline z ) are:
A. 3 and 2.
B. 3 and 2i.
C. 1 and 6.
D. 1 and 6i.
Verse 13. The solution of the system of equations (left{ begin{array}{l}x + 2y = 1 + i3x + iy = 2 – 3iend{array} right.) is:
A. (left{ begin{array}{l}x = 1 + iy = iend{array} right.).
B. (left{ begin{array}{l}x = iy = 1 + iend{array} right.).
C. (left{ begin{array}{l}x = 1 – iy = iend{array} right.).
D. (left{ begin{array}{l}x = iy = 1 – iend{array} right.).
Verse 14. Find a complex number with a real part of 12 and a modulus of 13.
A. (5 pm 12i).
B. 12 + 5i.
C. (12 pm 5i).
D. (12 pm i).
Verse 15. The equation ({z^2} – 2z + 3 = 0) has the following solutions:
A. (2 pm 2sqrt 2 i).
B. ( – 2 pm 2sqrt 2 i).
C. ( – 1 pm 2sqrt 2 i).
D. (1 pm sqrt 2 i).
Verse 16. The set of points representing complex numbers z satisfying (|overline z + 3 – 2i| = 4) is:
A. A circle with center I(3 ; 2) has radius R = 4.
B. The circle with center I(3 ; -2) has radius R= 4.
C. The circle with center I(-3 ; 2) has radius R = 4.
D. The circle with center I(- 3; -2) has radius R = 4.
Verse 17. Two points representing two conjugate complex numbers (z = 2 + 2i,,,overline z = 2 – 2i) are symmetric through :
A. Vertical axis.
B. Horizontal axis.
C. Origin coordinates.
D. Grade A(2; -2).
Verse 18. Given a complex number (z = rleft( {cos dfrac{pi }{2} + isin dfrac{pi }{2}} right)). Choose 1 accumen of z:
A. ( – dfrac{pi }{2}) B. ( – dfrac{{3pi }}{2})
C. (dfrac{{3pi }}{2}) D. (pi ).
Verse 19. Modulus of the sum of two complex numbers ({z_1} = 3 – 4i,,,,{z_2} = 4 + 3i):
A. (5sqrt 2 ) B. 10
C. 8 D. 50.
Verse 20. Given a complex number (z = – rleft( {cos varphi + isin varphi } right)). Find an accumen of z ?
A. ( – varphi ).
B. (varphi + 2pi ).
C. (varphi – 2pi ).
D. (varphi + pi ).
Verse 21. Calculate (z = dfrac{{5 + 5i}}{{3 – 4i}} + dfrac{{20}}{{4 + 3i}}).
A. z = 3 – i.
B. z = 3 + i.
C. z = – 3 – i.
D. z = – 3 + i.
Question 22.The set of points representing complex numbers z satisfying (|z + 1 + i|, le 2) is;
A. A circle with center I(1 ; 1) and radius R = 2.
B. A circle with center I(1; 1) and radius R = 2.
C. Circle with center I(- 1 ; – 1) radius R = 2.
D. Circle with center I(- 1 ; – 1) radius R = 2.
Verse 23. The trigonometric form of the complex number z = i – 1 is:
A. (z = sqrt 2 left( {cos dfrac{{3pi }}{4} – isin dfrac{{3pi }}{4}} right)).
B. (z = 2left( {cos dfrac{{3pi }}{4} + isin dfrac{{3pi }}{4}} right)).
C. (z = sqrt 2 left( {cos dfrac{{ – pi }}{4} + isin dfrac{{ – pi }}{4}} right)).
D. (z = sqrt 2 left( {cos dfrac{{3pi }}{4} + isin dfrac{{3pi }}{4}} right)).
Verse 24. In the complex plane, the points A, B are the representation points of ({z_1} = 2 – 4i,,,,{z_2} = 4 + 5i, respectively). The midpoint of AB has coordinates:
A. (Aleft( {3;dfrac{3}{2}} right)).
B. (Aleft( {3;1} right)).
C. (Aleft( {3;dfrac{1}{2}} right)).
D. (Aleft( {6;1} right)).
Verse 25. Given a complex number z such that (left( {3 + 2i} right)z + {left( {2 – i} right)^2} = 4 + i). The modulus of the complex number (w = left( {z + 1} right)overline z ) is:
A. 2 B. 4
C. 10 D. (sqrt {10} ).
Detailed explanation
first | 2 | 3 | 4 | 5 |
EASY | REMOVE | A | EASY | OLD |
6 | 7 | 8 | 9 | ten |
REMOVE | EASY | A | EASY | OLD |
11 | twelfth | 13 | 14 | 15 |
OLD | A | OLD | OLD | EASY |
16 | 17 | 18 | 19 | 20 |
A | REMOVE | REMOVE | A | EASY |
21 | 22 | 23 | 24 | 25 |
A | EASY | EASY | OLD | EASY |
Detailed explanation
Question 1: EASY
(begin{array}{l}{z_1} = 9 – i;{z_2} = – 3 + 2idfrac{{{z_1}}}{{{z_2}}} = dfrac{{ left( {9 – i} right)left( { – 3 – 2i} right)}}{{9 – 4{i^2}}},,,,,, , = dfrac{{ – 27 + 2{i^2} – 15i}}{{13}} = – dfrac{{29}}{{13}} – dfrac{{15}}{{13 }}i Rightarrow left| {dfrac{{{z_1}}}{{{z_2}}}} right| = sqrt {{{left( { – dfrac{{29}}{ {13}}} right)}^2} + {{left( { – dfrac{{15}}{{13}}} right)}^2}};;; ;;;;;;;;;, = sqrt {dfrac{{82}}{{13}}} end{array})
Verse 2: REMOVE
Question 3: A
(begin{array}{l}z = – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i Rightarrow {left( {overline z } right )^2} = {left( { – dfrac{1}{2} – dfrac{{sqrt 3 }}{2}i} right)^2},,,, ,,,,,,,,,,,,, = dfrac{{{{left( {1 + sqrt 3 i} right)}^2 }}}{4},,,,,,,,,,,,,,,,, = dfrac{{ – 2 + 2sqrt 3 i}}{4},,,,,,,,,,,,,,,,, = – dfrac {1}{2} + dfrac{{sqrt 3 }}{2}iend{array})
Question 4: EASY
Verse 5 C
(begin{array}{l}dfrac{{2 + i}}{{1 – i}}z = dfrac{{ – 1 + 3i}}{{2 + i}} Leftrightarrow z = dfrac{{left( { – 1 + 3i} right)left( {1 – i} right)}}{{{{(2 + i)}^2}}} Leftrightarrow z = dfrac{{2 + 4i}}{{3 + 4i}} Leftrightarrow z = dfrac{{left( {2 + 4i} right)left( {3 – 4i} right)} }{{9 – 16{i^2}}} Leftrightarrow z = dfrac{{6 – 16{i^2} + 4i}}{{25}} Leftrightarrow z = dfrac{{ 22}}{{25}} + dfrac{4}{{25}}i Rightarrow left| z right| = dfrac{{2sqrt 5 }}{5}end{array} )
Sentence 6: REMOVE
(begin{array}{l}z = {(1 + i)^{15}} = {left( {1 + i} right)^{14}}(1 + i), ,,, = {({(1 + + i)^2})^7}left( {1 + i} right) = {2^7}{i^7}left( {1 + i} right),,,, = – {2^7}ileft( {1 + i} right) = 128 – 128iend{array})
Question 7: EASY
Question 8: A
({z_1} – {z_2} = left( {2 – 5i} right) – ( – 2 – 3i))(, = 4 – 2i)
( Rightarrow left| {{z_1} – {z_2}} right| = 2sqrt 5)
Question 9: EASY
(begin{array}{l}left( {3 – 2i} right)z = 4 + 2i Leftrightarrow z = dfrac{{4 + 2i}}{{3 – 2i}} Leftrightarrow z = dfrac{{(4 + 2i)(3 + 2i)}}{{9 – 4{i^2}}},,,,,,,, ,,,, = dfrac{{12 + 4{i^2} + 14i}}{{13}},,,,,,,,, ,,, = dfrac{8}{{13}} + dfrac{{14}}{{13}}i Rightarrow overline z = dfrac{8}{{13}} – dfrac{{14}}{{13}}iend{array})
Question 10:
(begin{array}{l}{z^2} – 6z + 11 = 0 Leftrightarrow left( {{z^2} – 6z + 9} right) + 2 = 0 Leftrightarrow {(z – 3)^2} + 2 = 0 Rightarrow left[begin{array}{l}z–3=isqrt2z–3=–isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=3+isqrt2z=3–isqrt2end{array}rightend{array})[begin{array}{l}z–3=isqrt2z–3= –isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=3+isqrt2z=3–isqrt2end{array}rightend{array})
Question 11:
Question 12: A
({rm{w}} = 2z + overline z = 2(1 + 2i) + (1 – 2i) )(,= 3 + 2i)
real part: 3 , imaginary part: 2
Verse 13:
(left{ begin{array}{l}x + 2y = 1 + i3x + iy = 2 – 3iend{array} right. )
(Leftrightarrow left{ begin{array}{l}x = 1 + i – 2y{rm{ (1)}}3x + iy = 2 – 3i{rm{ (2)}} end{array} right.)
Substituting (1) into (2) we get:
(begin{array}{l}3(1 + i – 2y) + iy = 2 – 3i Leftrightarrow ( – 6 + i)y = – 1 – 6i Leftrightarrow y = dfrac{{ – 1 – 6i}}{{ – 6 + i}} Leftrightarrow y = dfrac{{left( { – 1 – 6i} right)left( { – 6 – i} right)}} {{36 – {i^2}}} = iend{array})
Replace y = i in (1) ( Rightarrow x = 1 – i)
Verse 14:
With the real part equal to 12, the complex number z has the form (z = 12 + bi)
(begin{array}{l}left| z right| = 13 Rightarrow left| {12 + bi} right| = 13 Leftrightarrow sqrt {{{12}^2} + { b^2}} = 13 Leftrightarrow {b^2} = 25 Leftrightarrow left[begin{array}{l}b=5Rightarrowz=12+5ib=–5Rightarrowz=12–5iend{array}rightend{array})[begin{array}{l}b=5Rightarrowz=12+5ib= –5Rightarrowz=12–5iend{array}rightend{array})
Question 15: EASY
(begin{array}{l}{z^2} – 2z + 3 = 0 Leftrightarrow left( {{z^2} – 2z + 1} right) + 2 = 0 Leftrightarrow {left( {z – 1} right)^2} + 2 = 0 Leftrightarrow {left( {z – 1} right)^2} = – 2 Rightarrow left[begin{array}{l}z–1=isqrt2z–1=–isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=1+isqrt2z=1–isqrt2end{array}rightend{array})[begin{array}{l}z–1=isqrt2z–1= –isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=1+isqrt2z=1–isqrt2end{array}rightend{array})
Verse 16: A
Verse 17: REMOVE
Verse 18: REMOVE
Question 19: A
(begin{array}{l}{z_1} + {z_2} = 3 – 4i + 4 + 3i = 7 – i Rightarrow left| {{z_1} + {z_2}} right| = 5 sqrt 2 end{array})
Question 20: EASY
Question 21: A
(begin{array}{l}z = dfrac{{5 + 5i}}{{3 – 4i}} + dfrac{{20}}{{4 + 3i}},, ,, = dfrac{{5left( {1 + i} right)left( {3 + 4i} right)}}{{9 – 16{i^2}}} + dfrac{{ 20left( {4 – 3i} right)}}{{16 – 9{i^2}}},,,, = dfrac{{5(3 + 4{i^2 } + 7i) + 20(4 – 3i)}}{{25}},,,, = dfrac{{5( – 1 + 7i) + 20left( {4 – 3i} right)}}{{25}} = 3 – iend{array})
Question 22: EASY
Set (z= x+yi)
(begin{array}{l}left| {z + 1 + i} right| le 2 Rightarrow left| {x + yi + 1 + i} right| le 2 Leftrightarrow left| {left( {x + 1} right) + left( {y + 1} right)} right| le 2 Leftrightarrow sqrt {{{left( {x) + 1} right)}^2} + {{left( {y + 1} right)}^2}} le 2end{array})
So the set of points representing complex numbers z is a circle with center I(-1, -1), radius equal to 2
Question 23: EASY
Verse 24:
Question 25: EASY
(begin{array}{l}left( {3 + 2i} right)z + {left( {2 – i} right)^2} = 4 + i Leftrightarrow left( { 3 + 2i} right)z + (3 – 4i) = 4 + i Leftrightarrow left( {3 + 2i} right)z = 1 + 5i Leftrightarrow z = dfrac{{1 + 5i}}{{3 + 2i}} Leftrightarrow z = dfrac{{left( {1 + 5i} right)left( {3 – 2i} right)}}{{9 – 4{ i^2}}} Leftrightarrow z = dfrac{{13 + 13i}}{{13}} = 1 + i{rm{w}} = (z + 1)overline z = ( 2 + i)(1 – i),,,,,, = 2 – {i^2} – i = 3 – i Rightarrow left| {rm{w} } right| = sqrt {10} end{array})
[rule_{ruleNumber}]
[box type=”note” align=”” class=”” s14 lineheight”>Answers and detailed solutions 45-minute test (1 period) – Question 4 – Chapter IV – Calculus 12.
Topic
Question 1. Given two complex numbers ({z_1} = 9 – i,,,,{z_2} = – 3 + 2i). Calculate the value of (left| {dfrac{{{z_1}}}{{{z_2}}}} right|) equal to /
A. (dfrac{{2sqrt {154} }}{{13}}). B. (dfrac{{616}}{{169}}).
C. (dfrac{{82}}{{13}}). D. (sqrt {dfrac{{82}}{{13}}} ).
Verse 2. Given two complex numbers ({z_1} = a + bi,,,{z_2} = c + di)z. Find the real part of the complex number ({z_1}. {z_2}).
A. The real part of the complex number ({z_1}.{z_2}) is ac + bd.
B. The real part of the complex number ({z_1}. {z_2}) is ac – bd .
C. The real part of the complex number ({z_1}.{z_2}) is ad + bc.
D. The real part of the complex number ({z_1}.{z_2}) is ad – bc
Verse 3. Given a complex number (z = – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i). Then the complex number ({left( {overline z } right)^2}) is equal to ;
A. ( – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i).
B. (sqrt 3 – i).
C. ( – dfrac{1}{2} – dfrac{{sqrt 3 }}{2}i).
D. (1 + sqrt 3 i).
Verse 4.Assume that A, B are respectively the representation of complex numbers ({z_1} = {a_1} + {b_1}i,,,,{z_2} = {a_2} + {b_2}i ). Then the length of the vector (overrightarrow {AB} ) is equal to ;
A. (|{z_1} + {z_2}|).
B. (|{z_1}| + |{z_2}|).
C. (|{z_1}| – |{z_2}|).
D. (|{z_1} – {z_2}|).
Question 5. The modulus of the complex number z satisfying (dfrac{{2 + i}}{{1 – i}}z = dfrac{{ – 1 + 3i}}{{2 + i}}) is:
A. (sqrt 5 ) B. (dfrac{{sqrt 5 }}{5})
C. (dfrac{{2sqrt 5 }}{5}) D. (dfrac{{3sqrt 5 }}{5}).
Verse 6. Calculate the following complex number: (z = {left( {1 + i} right)^{15}}).
A. (z = – 128 + 128i).
B. (z = 128 – 128i).
C. (z = 128 + 128i).
D. (z = – 128 – 128i).
Verse 7. Let the complex number z = a + bi. Then the number (dfrac{1}{2}left( {z + overline z } right)) is:
A. A pure imaginary number.
B. 2a.
C. i.
D. a.
Verse 8. For complex numbers ({z_1} = 2 – 5i,,,,{z_2} = – 2 – 3i). Let’s calculate (|{z_1} – {z_2}|).
A. (2sqrt 5 ) B. 20
C. 12 D. (2sqrt 3 ).
Verse 9. Given a complex number z satisfying (left( {3 – 2i} right)z = 4 + 2i). Find the conjugate complex number of z.
A. (overline z = 4 – 2i).
B. (overline z = dfrac{8}{{13}} + dfrac{{14}}{{13}}i).
C. (overline z = 3 + 2i).
D. (overline z = dfrac{8}{{13}} – dfrac{{14}}{{13}}i).
Question 10. Solving the equation ({z^2} – 6z + 11 = 0), we have a solution of :
A. (z = 3 + sqrt 2 i).
B. (z = 3 – sqrt 2 i).
C. (left[begin{array}{l}z=3+sqrt2iz=3–sqrt2iend{array}right)[begin{array}{l}z=3+sqrt2iz=3–sqrt2iend{array}right)
D. Another result.
Verse 11. Given two complex numbers (z = a + bi,,,,z’ = a’ + b’i). Choose the correct formula.
A. (z + z’ = left( {a + b} right) + left( {a’ + b’} right)i).
B. (z – z’ = left( {a + a’} right) – left( {b + b’} right)i).
C. (z.z’ = left( {aa’ – bb’} right) + left( {ab’ + a’b} right)i).
D. (z.z’ = left( {aa’ + bb’} right) – left( {ab’ + a’b} right)i).
Verse 12. Let z = 1 + 2i. The real and imaginary parts of the complex number (w = 2z + overline z ) are:
A. 3 and 2.
B. 3 and 2i.
C. 1 and 6.
D. 1 and 6i.
Verse 13. The solution of the system of equations (left{ begin{array}{l}x + 2y = 1 + i3x + iy = 2 – 3iend{array} right.) is:
A. (left{ begin{array}{l}x = 1 + iy = iend{array} right.).
B. (left{ begin{array}{l}x = iy = 1 + iend{array} right.).
C. (left{ begin{array}{l}x = 1 – iy = iend{array} right.).
D. (left{ begin{array}{l}x = iy = 1 – iend{array} right.).
Verse 14. Find a complex number with a real part of 12 and a modulus of 13.
A. (5 pm 12i).
B. 12 + 5i.
C. (12 pm 5i).
D. (12 pm i).
Verse 15. The equation ({z^2} – 2z + 3 = 0) has the following solutions:
A. (2 pm 2sqrt 2 i).
B. ( – 2 pm 2sqrt 2 i).
C. ( – 1 pm 2sqrt 2 i).
D. (1 pm sqrt 2 i).
Verse 16. The set of points representing complex numbers z satisfying (|overline z + 3 – 2i| = 4) is:
A. A circle with center I(3 ; 2) has radius R = 4.
B. The circle with center I(3 ; -2) has radius R= 4.
C. The circle with center I(-3 ; 2) has radius R = 4.
D. The circle with center I(- 3; -2) has radius R = 4.
Verse 17. Two points representing two conjugate complex numbers (z = 2 + 2i,,,overline z = 2 – 2i) are symmetric through :
A. Vertical axis.
B. Horizontal axis.
C. Origin coordinates.
D. Grade A(2; -2).
Verse 18. Given a complex number (z = rleft( {cos dfrac{pi }{2} + isin dfrac{pi }{2}} right)). Choose 1 accumen of z:
A. ( – dfrac{pi }{2}) B. ( – dfrac{{3pi }}{2})
C. (dfrac{{3pi }}{2}) D. (pi ).
Verse 19. Modulus of the sum of two complex numbers ({z_1} = 3 – 4i,,,,{z_2} = 4 + 3i):
A. (5sqrt 2 ) B. 10
C. 8 D. 50.
Verse 20. Given a complex number (z = – rleft( {cos varphi + isin varphi } right)). Find an accumen of z ?
A. ( – varphi ).
B. (varphi + 2pi ).
C. (varphi – 2pi ).
D. (varphi + pi ).
Verse 21. Calculate (z = dfrac{{5 + 5i}}{{3 – 4i}} + dfrac{{20}}{{4 + 3i}}).
A. z = 3 – i.
B. z = 3 + i.
C. z = – 3 – i.
D. z = – 3 + i.
Question 22.The set of points representing complex numbers z satisfying (|z + 1 + i|, le 2) is;
A. A circle with center I(1 ; 1) and radius R = 2.
B. A circle with center I(1; 1) and radius R = 2.
C. Circle with center I(- 1 ; – 1) radius R = 2.
D. Circle with center I(- 1 ; – 1) radius R = 2.
Verse 23. The trigonometric form of the complex number z = i – 1 is:
A. (z = sqrt 2 left( {cos dfrac{{3pi }}{4} – isin dfrac{{3pi }}{4}} right)).
B. (z = 2left( {cos dfrac{{3pi }}{4} + isin dfrac{{3pi }}{4}} right)).
C. (z = sqrt 2 left( {cos dfrac{{ – pi }}{4} + isin dfrac{{ – pi }}{4}} right)).
D. (z = sqrt 2 left( {cos dfrac{{3pi }}{4} + isin dfrac{{3pi }}{4}} right)).
Verse 24. In the complex plane, the points A, B are the representation points of ({z_1} = 2 – 4i,,,,{z_2} = 4 + 5i, respectively). The midpoint of AB has coordinates:
A. (Aleft( {3;dfrac{3}{2}} right)).
B. (Aleft( {3;1} right)).
C. (Aleft( {3;dfrac{1}{2}} right)).
D. (Aleft( {6;1} right)).
Verse 25. Given a complex number z such that (left( {3 + 2i} right)z + {left( {2 – i} right)^2} = 4 + i). The modulus of the complex number (w = left( {z + 1} right)overline z ) is:
A. 2 B. 4
C. 10 D. (sqrt {10} ).
Detailed explanation
first | 2 | 3 | 4 | 5 |
EASY | REMOVE | A | EASY | OLD |
6 | 7 | 8 | 9 | ten |
REMOVE | EASY | A | EASY | OLD |
11 | twelfth | 13 | 14 | 15 |
OLD | A | OLD | OLD | EASY |
16 | 17 | 18 | 19 | 20 |
A | REMOVE | REMOVE | A | EASY |
21 | 22 | 23 | 24 | 25 |
A | EASY | EASY | OLD | EASY |
Detailed explanation
Question 1: EASY
(begin{array}{l}{z_1} = 9 – i;{z_2} = – 3 + 2idfrac{{{z_1}}}{{{z_2}}} = dfrac{{ left( {9 – i} right)left( { – 3 – 2i} right)}}{{9 – 4{i^2}}},,,,,, , = dfrac{{ – 27 + 2{i^2} – 15i}}{{13}} = – dfrac{{29}}{{13}} – dfrac{{15}}{{13 }}i Rightarrow left| {dfrac{{{z_1}}}{{{z_2}}}} right| = sqrt {{{left( { – dfrac{{29}}{ {13}}} right)}^2} + {{left( { – dfrac{{15}}{{13}}} right)}^2}};;; ;;;;;;;;;, = sqrt {dfrac{{82}}{{13}}} end{array})
Verse 2: REMOVE
Question 3: A
(begin{array}{l}z = – dfrac{1}{2} + dfrac{{sqrt 3 }}{2}i Rightarrow {left( {overline z } right )^2} = {left( { – dfrac{1}{2} – dfrac{{sqrt 3 }}{2}i} right)^2},,,, ,,,,,,,,,,,,, = dfrac{{{{left( {1 + sqrt 3 i} right)}^2 }}}{4},,,,,,,,,,,,,,,,, = dfrac{{ – 2 + 2sqrt 3 i}}{4},,,,,,,,,,,,,,,,, = – dfrac {1}{2} + dfrac{{sqrt 3 }}{2}iend{array})
Question 4: EASY
Verse 5 C
(begin{array}{l}dfrac{{2 + i}}{{1 – i}}z = dfrac{{ – 1 + 3i}}{{2 + i}} Leftrightarrow z = dfrac{{left( { – 1 + 3i} right)left( {1 – i} right)}}{{{{(2 + i)}^2}}} Leftrightarrow z = dfrac{{2 + 4i}}{{3 + 4i}} Leftrightarrow z = dfrac{{left( {2 + 4i} right)left( {3 – 4i} right)} }{{9 – 16{i^2}}} Leftrightarrow z = dfrac{{6 – 16{i^2} + 4i}}{{25}} Leftrightarrow z = dfrac{{ 22}}{{25}} + dfrac{4}{{25}}i Rightarrow left| z right| = dfrac{{2sqrt 5 }}{5}end{array} )
Sentence 6: REMOVE
(begin{array}{l}z = {(1 + i)^{15}} = {left( {1 + i} right)^{14}}(1 + i), ,,, = {({(1 + + i)^2})^7}left( {1 + i} right) = {2^7}{i^7}left( {1 + i} right),,,, = – {2^7}ileft( {1 + i} right) = 128 – 128iend{array})
Question 7: EASY
Question 8: A
({z_1} – {z_2} = left( {2 – 5i} right) – ( – 2 – 3i))(, = 4 – 2i)
( Rightarrow left| {{z_1} – {z_2}} right| = 2sqrt 5)
Question 9: EASY
(begin{array}{l}left( {3 – 2i} right)z = 4 + 2i Leftrightarrow z = dfrac{{4 + 2i}}{{3 – 2i}} Leftrightarrow z = dfrac{{(4 + 2i)(3 + 2i)}}{{9 – 4{i^2}}},,,,,,,, ,,,, = dfrac{{12 + 4{i^2} + 14i}}{{13}},,,,,,,,, ,,, = dfrac{8}{{13}} + dfrac{{14}}{{13}}i Rightarrow overline z = dfrac{8}{{13}} – dfrac{{14}}{{13}}iend{array})
Question 10:
(begin{array}{l}{z^2} – 6z + 11 = 0 Leftrightarrow left( {{z^2} – 6z + 9} right) + 2 = 0 Leftrightarrow {(z – 3)^2} + 2 = 0 Rightarrow left[begin{array}{l}z–3=isqrt2z–3=–isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=3+isqrt2z=3–isqrt2end{array}rightend{array})[begin{array}{l}z–3=isqrt2z–3= –isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=3+isqrt2z=3–isqrt2end{array}rightend{array})
Question 11:
Question 12: A
({rm{w}} = 2z + overline z = 2(1 + 2i) + (1 – 2i) )(,= 3 + 2i)
real part: 3 , imaginary part: 2
Verse 13:
(left{ begin{array}{l}x + 2y = 1 + i3x + iy = 2 – 3iend{array} right. )
(Leftrightarrow left{ begin{array}{l}x = 1 + i – 2y{rm{ (1)}}3x + iy = 2 – 3i{rm{ (2)}} end{array} right.)
Substituting (1) into (2) we get:
(begin{array}{l}3(1 + i – 2y) + iy = 2 – 3i Leftrightarrow ( – 6 + i)y = – 1 – 6i Leftrightarrow y = dfrac{{ – 1 – 6i}}{{ – 6 + i}} Leftrightarrow y = dfrac{{left( { – 1 – 6i} right)left( { – 6 – i} right)}} {{36 – {i^2}}} = iend{array})
Replace y = i in (1) ( Rightarrow x = 1 – i)
Verse 14:
With the real part equal to 12, the complex number z has the form (z = 12 + bi)
(begin{array}{l}left| z right| = 13 Rightarrow left| {12 + bi} right| = 13 Leftrightarrow sqrt {{{12}^2} + { b^2}} = 13 Leftrightarrow {b^2} = 25 Leftrightarrow left[begin{array}{l}b=5Rightarrowz=12+5ib=–5Rightarrowz=12–5iend{array}rightend{array})[begin{array}{l}b=5Rightarrowz=12+5ib= –5Rightarrowz=12–5iend{array}rightend{array})
Question 15: EASY
(begin{array}{l}{z^2} – 2z + 3 = 0 Leftrightarrow left( {{z^2} – 2z + 1} right) + 2 = 0 Leftrightarrow {left( {z – 1} right)^2} + 2 = 0 Leftrightarrow {left( {z – 1} right)^2} = – 2 Rightarrow left[begin{array}{l}z–1=isqrt2z–1=–isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=1+isqrt2z=1–isqrt2end{array}rightend{array})[begin{array}{l}z–1=isqrt2z–1= –isqrt2end{array}rightLeftrightarrowleft[begin{array}{l}z=1+isqrt2z=1–isqrt2end{array}rightend{array})
Verse 16: A
Verse 17: REMOVE
Verse 18: REMOVE
Question 19: A
(begin{array}{l}{z_1} + {z_2} = 3 – 4i + 4 + 3i = 7 – i Rightarrow left| {{z_1} + {z_2}} right| = 5 sqrt 2 end{array})
Question 20: EASY
Question 21: A
(begin{array}{l}z = dfrac{{5 + 5i}}{{3 – 4i}} + dfrac{{20}}{{4 + 3i}},, ,, = dfrac{{5left( {1 + i} right)left( {3 + 4i} right)}}{{9 – 16{i^2}}} + dfrac{{ 20left( {4 – 3i} right)}}{{16 – 9{i^2}}},,,, = dfrac{{5(3 + 4{i^2 } + 7i) + 20(4 – 3i)}}{{25}},,,, = dfrac{{5( – 1 + 7i) + 20left( {4 – 3i} right)}}{{25}} = 3 – iend{array})
Question 22: EASY
Set (z= x+yi)
(begin{array}{l}left| {z + 1 + i} right| le 2 Rightarrow left| {x + yi + 1 + i} right| le 2 Leftrightarrow left| {left( {x + 1} right) + left( {y + 1} right)} right| le 2 Leftrightarrow sqrt {{{left( {x) + 1} right)}^2} + {{left( {y + 1} right)}^2}} le 2end{array})
So the set of points representing complex numbers z is a circle with center I(-1, -1), radius equal to 2
Question 23: EASY
Verse 24:
Question 25: EASY
(begin{array}{l}left( {3 + 2i} right)z + {left( {2 – i} right)^2} = 4 + i Leftrightarrow left( { 3 + 2i} right)z + (3 – 4i) = 4 + i Leftrightarrow left( {3 + 2i} right)z = 1 + 5i Leftrightarrow z = dfrac{{1 + 5i}}{{3 + 2i}} Leftrightarrow z = dfrac{{left( {1 + 5i} right)left( {3 – 2i} right)}}{{9 – 4{ i^2}}} Leftrightarrow z = dfrac{{13 + 13i}}{{13}} = 1 + i{rm{w}} = (z + 1)overline z = ( 2 + i)(1 – i),,,,,, = 2 – {i^2} – i = 3 – i Rightarrow left| {rm{w} } right| = sqrt {10} end{array})
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