Solve Q1, 2, 3 explaining detailly each step COMPLEX NUMBERSa) MCQS:1. (2+3i)(1 – i) =2. i =A. 2 – 2iB. 2+ 4i C. 5 + i D. – 1+ i%3DA. -1 B. -iС. 1D. i3. The imaginary part of i"(2 + V3i) – i(1 – 2i) is: A. V3 –1 B. 0 C. - V3 -1 D. - 44. The complex conjugate of (1- i) + (2i - 3)i is:A. -2 -i B. 3 + 2i5. (V3 + i) + (1 – V3i) =C. 3 – 2i D. -1-2iА. 1B. -iC. i D. -1D. +21.A. - -- L52. 1iC. .=+3316.2+i2B.-3 35 57. If (3 – 2i)z - (7 + 4i) = 0, then z =A. +132629B.+ 2i292C. 1+ 2i D.--_re1313-1+iv38.-1-iv3A. -1- iv3B. -(2 + iv/3)C. - (1+ iv3)D. –(1+ iv3)|-229. Given that z2--1+3i, z+A. -- B- C-- D-+124В.184C.-1271155310. Given that Z1= 1 -i and z = 2 + i, z;-=Z.24A.524.B.523.C.52i58.+ - iD.5511. Given that z and Z are conjugate complex numbers, which cne of the following is not alwaystrue?A. Re (z) = Re(2) B. Im (z) + Im(2) = 0 C. Re(2) = Im(z) D. (z)12. The roots cf the equation: z- 6z+ 34 = 0 areA. -2, 8 B. 6- 10i, 6 + 10i C. -4, 16 D. 3 - 5i, 3 + 5i13. The quadratic equation with 2 -i as one of its roots is:A. x+ 5x – 4 0 B. x- 5x + 4= 0 C. x - 4x + 4 = 014. p+ 2 is a root of x-x+ 0. The values of p and q are:(z)D. x² - 4x – 5 = (0%3D17A. p=B. p = 1, q = 54.117C. p15. Given that 5(a + ib) +3 - 2i = 6i, the values of a and b are:D. p =, G = "172A. a = -3, b -8B. a =8.5'3C. a =, b = =3С.аa =, b =8.512116. If(x + iy), the values of x and y are:|3-i1+2i107.A. x =B. x = 7, y= 7 C. x 7, y = -7 D. x = -7, y = 7101017. The complex number -2 + 2i can also be expressed asA. 2 [cos (-) + i sin (-ĐIA. 2 cos (-)+ i sinB. 2 cos () + i sin+ i sin () D. 2V2 cos (-) + i sin (-|TTOSC. 2v2 fcos () + i sin ( D.2V2 (cos (-) + isin(-)3T44418. Find the modulus sroument form of the complex number: z = -V3 - iA. A. 2 [cos +i sm B. z [cos-1sinc 2 cost -i sinTT[ca:COSCOS+ i sinD. 2 cos-i sin5TT74IN

The solution is given as follows

COMPLEX NUMBERS a) MCQS: 1. (2 + 3i)(1 – i) = 2. i = 3. The imaginary part of i (2 + V3i) – i(1 - 2i) is: A. V3 – 1 A. 2 – 2i B. 2+ 4i C. 5 + i D. - 1+i A. -1 B. -i С. 1 D. i B. 0 C. - V3 -1 D. - 4 4. The complex conjugate of (1 - i) + (2i - 3)i is: A. -2- i B. 3 + 2i 5. (V3 + i) (1 – V3i) = С. 3 - 2i А. 1 D. -1-2i C. i В. -і 2. 1 C. .+i D. -1 6. 2+i A. 2 -i 2 В. 1 i 3 2 D. 1 i 3. 3 7. If (3 – 2i)z -(7+4i) = 0, then z= A. + 2 B. + 2i 13 C. I + 2i D. -+ В. + 2i i 13 5 3. 13 -1+iV3 8. -1-iV3 B. - (2 + iv3) ) D.-(1 + iv3) A. -1- iv3 C.(1+ iv3) 9. Given that z = =-1+3i, z+ =

COMPLEX NUMBERS a) MCQS: 1. (2 + 3i)(1 – i) = 2. i = 3. The imaginary part of i (2 + V3i) – i(1 - 2i) is: A. V3 – 1 A. 2 – 2i B. 2+ 4i C. 5 + i D. - 1+i A. -1 B. -i С. 1 D. i B. 0 C. - V3 -1 D. - 4 4. The complex conjugate of (1 - i) + (2i - 3)i is: A. -2- i B. 3 + 2i 5. (V3 + i) (1 – V3i) = С. 3 - 2i А. 1 D. -1-2i C. i В. -і 2. 1 C. .+i D. -1 6. 2+i A. 2 -i 2 В. 1 i 3 2 D. 1 i 3. 3 7. If (3 – 2i)z -(7+4i) = 0, then z= A. + 2 B. + 2i 13 C. I + 2i D. -+ В. + 2i i 13 5 3. 13 -1+iV3 8. -1-iV3 B. - (2 + iv3) ) D.-(1 + iv3) A. -1- iv3 C.(1+ iv3) 9. Given that z = =-1+3i, z+ =

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Question

Solve Q1, 2, 3 explaining detailly each step

COMPLEX NUMBERS
a) MCQS:
1. (2+3i)(1 – i) =
2. i =
A. 2 – 2i
B. 2+ 4i C. 5 + i D. – 1+ i
%3D
A. -1 B. -i
С. 1
D. i
3. The imaginary part of i"(2 + V3i) – i(1 – 2i) is: A. V3 –1 B. 0 C. - V3 -1 D. - 4
4. The complex conjugate of (1- i) + (2i - 3)i is:
A. -2 -i B. 3 + 2i
5. (V3 + i) + (1 – V3i) =
C. 3 – 2i D. -1-2i
А. 1
B. -i
C. i D. -1
D. +
2
1.
A. - -- L
5
2. 1
i
C. .=+
3
3
1
6.
2+i
2
B.
-
3 3
5 5
7. If (3 – 2i)z - (7 + 4i) = 0, then z =
A. +
13
26
29
B.+ 2i
29
2
C. 1+ 2i D.
-
-_re
13
13
-1+iv3
8.
-1-iv3
A. -1- iv3
B. -(2 + iv/3)
C. - (1+ iv3)
D. –(1+ iv3)
|
-
2
2
9. Given that z
2
--1+3i, z+
A. -- B- C-- D-+
12
4
В.
18
4
C.-
12
7
11
5
5
3
10. Given that Z1
= 1 -i and z = 2 + i, z;-=
Z.2
4
A.
5
2
4.
B.
5
2
3.
C.
5
2
i
5
8.
+ - i
D.
5
5
11. Given that z and Z are conjugate complex numbers, which cne of the following is not always
true?
A. Re (z) = Re(2) B. Im (z) + Im(2) = 0 C. Re(2) = Im(z) D. (z)
12. The roots cf the equation: z- 6z+ 34 = 0 are
A. -2, 8 B. 6- 10i, 6 + 10i C. -4, 16 D. 3 - 5i, 3 + 5i
13. The quadratic equation with 2 -i as one of its roots is:
A. x+ 5x – 4 0 B. x- 5x + 4= 0 C. x - 4x + 4 = 0
14. p+ 2 is a root of x-x+ 0. The values of p and q are:
(z)
D. x² - 4x – 5 = (0
%3D
17
A. p=
B. p = 1, q = 5
4.
1
17
C. p
15. Given that 5(a + ib) +3 - 2i = 6i, the values of a and b are:
D. p =, G = "
17
2
A. a = -3, b -8
B. a =
8.
5'
3
C. a =, b = =
3
С.а
a =, b =
8.
5
1
2
1
16. If
(x + iy), the values of x and y are:
|
3-i
1+2i
10
7.
A. x =
B. x = 7, y= 7 C. x 7, y = -7 D. x = -7, y = 7
10
10
17. The complex number -2 + 2i can also be expressed as
A. 2 [cos (-) + i sin (-ĐI
A. 2 cos (-)+ i sin
B. 2 cos () + i sin
+ i sin () D. 2V2 cos (-) + i sin (-|
TT
OS
C. 2v2 fcos () + i sin ( D.2V2 (cos (-) + isin(-)
3T
4
4
4
18. Find the modulus sroument form of the complex number: z = -V3 - i
A. A. 2 [cos +i sm B. z [cos-1sinc 2 cost -i sin
TT
[ca:
COS
COS
+ i sinD. 2 cos-
i sin
5TT
74
IN

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