OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(b)

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S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(b)

Question 1.
In each of the following find r + s, r – s, rs, \(\frac { r }{ s }\) if r denotes the first complex number and s denotes the second complex number :
(i) 3 + 7i, i
(ii) – i, 5 + 2 i
(iii) 3i, 1 – i
(iv) – 7, – 1 – 3i
(v) 7 + 3i, 3i – 7
Solution:
(i) Given r = 3 + 7i and s = i
∴ r + s = 3 + 7i + i = 3 + 8i;
rs = (3 + 7i) i = 3i + 7i² = 3i – 7
r – s = 3 + 7i – i = 3 + 6i;
\(\frac{r}{s}=\frac{3+7 i}{i} \times \frac{i}{i}=\frac{3 i-7}{-1}\) = 7 – 3 i

(ii) Given r = – i; s = 5 + 2i
∴ r + s = – i + 5 + 2i = 5 + i;
r – s = – i – 5 – 2i = – 5 – 3i
r . s = – i (5 + 2i) = – 5i – 2i² = – 5i + 2
\(\frac{r}{s}=\frac{-i}{5+2 i} \times \frac{5-2 i}{5-2 i}=\frac{-i(5-2 i)}{5^2-(2 i)^2}\)
= \(\frac{-5 i-2}{25+4}=-\frac{5}{29} i-\frac{2}{29}\)

(iii) Given r = 3i and s = 1 – i
∴ r + s = 3i + 1 – i = 1 + 2i;
r – s = 3i – 1 + i = 4i – 1
rs = 3i (1 – i) = 3i – (- 3) = 3i + 3
\(\frac{r}{s}=\frac{3 i}{1-i} \times \frac{1+i}{1+i}=\frac{3 i(1+i)}{1^2-i^2}\)
= \(\frac{3 i+3 i^2}{1+1}=\frac{3}{2} i-\frac{3}{2}\)

(iv) Given r = – 7 and s = – 1 – 3i
∴ r + s = – 7 – 1 – 3i = – 8 – 3i;
r – s = – 7 + 1 + 3i = – 6 + 3i
rs = – 7 (- 1 – 3i) = 7 + 21i
and \(\frac{r}{s}=\frac{-7}{-1-3 i}=\frac{7}{3 i+1} \times \frac{3 i-1}{3 i-1}\)
= \(\frac{7(3 i-1)}{-9-1}=\frac{21 i-7}{-10}=\frac{-21}{10} i+\frac{7}{10}\)

(v) Given r = 7 + 3i; s = 3i – 7
∴ r + s = 7 + 3i + 3i – 7 = 6i;
r – s = 7 + 3i – 3i + 7 = 14
rs = (7 + 3i) (3i – 7) = (3i)² – 7²
= – 9 – 49 = – 58
\(\frac{r}{s}=\frac{7+3 i}{3 i-7} \times \frac{3 i+7}{3 i+7}=\frac{(3 i+7)^2}{-9-49}\)
= \(\frac{-9+49+42 i}{-58}=\frac{40+42 i}{-58}\)
= \(\frac{-20}{29}-\frac{21}{29} i\)

Question 2.
Solve each of the following equations for real x and y :
(i) (x + y) + (3 – 2i) = 1 + 4i
(ii) (x + yi) – (7 + 4i) = 3 – 5i
(iii) 2x + yi = 1 + (2 + 3i)
(iv) x + 2yi = i – (- 3 + 5i)
Solution:
(i) Given, (x + yi) + (3 – 2i) = 1 + 4i
⇒ (x + 3) + i (y – 2) = 1 + 4i
On comparing real and imaginary parts on both sides ; we have
x + 3 = 1 ⇒ x = – 2
y – 2 = 4 ⇒ y = 6

(ii) Given (x + yi) – (7 + 4i) = 3 – 5i
⇒ (x – 7) + i (y – 4) = 3 – 5i
On comparing real and imaginary parts on both sides ; we have
x – 7 = 3 ⇒ x = 10
and y – 4 = – 5 ⇒ y = – 1

(iii) Given 2x + yi = 1 + (2 + 3i)
⇒ 2x + yi = 3 + 3i
On comparing real and imaginary parts on both sides, we have
2x = 3 ⇒ x = \(\frac { 3 }{ 2 }\) and y = 3

(iv) Given x + 2yi = i – (- 3 + 5i) = i + 3 – 5i
⇒ x + 2yi = 3 – 4i
On comparing real and imaginary parts on both sides,
x = 3 and 2y = – 4 ⇒ y = – 2

Question 3.
Determine the conjugate and the reciprocal of each complex number given below :
(i) i
(ii) i³
(iii) 3 – i
(iv) \(\sqrt{-1}-3\)
(v) \(\sqrt{-9}-3\)
Solution:
(i) Conjugate of i = \(\bar{i}\) = – i
and Reciprocal of i = \(\frac{1}{i}=\frac{-i^2}{i}\)= – i

(ii) Conjugate of i³ = \(\bar{i}\)³ = (- i)³ = – i³
= – i² x i = i
and Reciprocal of i³ = \(\frac{1}{i^3}\)
= \(\frac{1}{-i} \times \frac{i}{i}=\frac{i}{-i^2}\) = i

(iii) Conjugate of (3 – i) = \(\overline{3-i}\) = 3 + i

Question 4.
Simplify :
(i) (3 – 7i)²
(ii) \(\left(\frac{-1}{2}-\frac{\sqrt{3}}{2} i\right)^2\)
(iii) (9 + 4i) (\(\frac { 3 }{ 2 }\) – i ) (9 – 4i)
Solution:
(i) (3 – 7i)² = 3² x (7i)² – 42i
= 9 – 49 – 42i
= – 40 – 42i

(ii) \(\left(\frac{-1}{2}-\frac{\sqrt{3}}{2} i\right)^2\)
= \(\left(-\frac{1}{2}\right)^2+\left(-\frac{\sqrt{3}}{2} i\right)^2+\frac{\sqrt{3}}{2} i\)
= \(\frac{1}{4}+\frac{3}{4} i^2+\frac{\sqrt{3}}{2} i=\frac{1}{4}-\frac{3}{4}+\frac{\sqrt{3}}{2} i\)
= \(\frac{-1}{2}+\frac{\sqrt{3}}{2} i\)

(iii) (9 + 4i)(\(\frac { 3 }{ 2 }\) – i)(9 – 4i)
= (9 + 4i)(9 – 4i)(\(\frac { 3 }{ 2 }\) – i)
= [81 – (4i)²](\(\frac { 3 }{ 2 }\) – i)
= (81 + 16) (\(\frac { 3 }{ 2 }\) – i)
= 97(\(\frac { 3 }{ 2 }\) – i) = \(\frac { 291 }{ 2 }\) – 97i

Question 5.
Determine real values of x and y for which each statement is true.
(i) \(\frac { x+y }{ i }\) + x – y + 4 = 0
(ii) – (x + 3y)i + (2x – y + 1) = \(\frac { 8 }{ i }\)
(iii) (x – yi) = \(\frac{2+i}{1+i}\)
(iv) (3 – 4i) + (x + yi) = 1 + 0i
(v) (x – yi)(2 + 3i) = \(\frac{x-2 i}{1-i}\)
(vi) (x⁴ + 2xi) – (3x² + yi) = (3 – 5i) + (1 + 2yi)
Solution:
(i) Given \(\frac { x+y }{ i }\) + x – y + 4 = 0
⇒ (x + y) + i (x – y + 4) = 0 = 0 + i0
On comparing real and imaginary parts on both sides ; we have
x + y = 0 …(1)
and x – y = – 4 …(2)
On adding eqn. (1) and (2); we have 2x = – 4
⇒ x = – 2 from (1) ; y = 2

(ii) Given – (x + 3y) i + (2x – y + 1) = \(\frac { 8 }{ i }\)
⇒ (x + 3y) + (2x – y + 1) i = 8 [∵ i² = – 1]
On comparing real and imaginary parts on both sides ; we have
x + 3y = 8 …(1)
2x – y = – 1 …(2)
eqn. (1) + 3 x eqn. (2); we have
7x = 5 ⇒ x = \(\frac { 5 }{ 7 }\)
∴ from (1); \(\frac { 5 }{ 7 }\) + 3y =8
⇒ 3y = 8 – \(\frac { 5 }{ 7 }\) = \(\frac { 51 }{ 7 }\) ⇒ y = \(\frac { 17 }{ 7 }\)

(iii) Given (x – yi) = \(\frac{2+i}{1+i} \times \frac{1-i}{1-i}\)
= \(\frac{(2+i)(1-i)}{1^2-i^2}\)
⇒ (x – yi) = \(\frac{2-2 i+i+1}{1+1}=\frac{3-i}{2}\)
= \(\frac { 3 }{ 2 }\) – \(\frac { i }{ 2 }\)
On comparing real and imaginary parts on both sides; we have
x = \(\frac { 3 }{ 2 }\) and – y = – \(\frac { 1 }{ 2 }\) ⇒ y = \(\frac { 1 }{ 2 }\)

(iv) Given (3 – 4i) (x + iy) = 1 + 0i
⇒ x + iy = \(\frac{1}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{3+4 i}{3^2-(4 i)^2}\)
⇒ x + iy = \(\frac{3+4 i}{9+16}=\frac{3}{25}+\frac{4}{25}\)i
On comparing real and imaginary parts on both sides ; we have
x = \(\frac { 3 }{ 25 }\) and y = \(\frac { 4 }{ 25 }\)

(v) Given (x – yi) (2 + 3i) = \(\frac{x-2 i}{1-i}\)
⇒ (x – yi) (2 + 30 (1 – i) = x – 2i
⇒ (x – yi) (2 – 2i + 3i + 3) = x – 2i
⇒ (x – yi) (5 + i) = x – 2i
⇒ 5x + xi – 5yi + y = x – 2i
On comparing real and imaginary parts on both sides ; we have
5x + y = x ⇒ 4x + y = 0 …(1)
x – 5y = – 2 …(2)
∴ from (1) ; y = -4x
∴ from (2); x + 20x = – 2 ⇒ x = \(\frac { -2 }{ 21 }\)
∴ from (1) ; y = – 4x = \(\frac { 8 }{ 21 }\)

(vi) Given
(x⁴ + 2xi) – (3x² + yi) = (3 – 5i) + 1 + 2yi
⇒ (x⁴ – 3x²) + i (2x – y) = 4 + (2y – 5) i
On comparing real and imaginary parts on both sides, we have
x⁴ – 3x² = 4
⇒ x⁴ – 3x² – 4 = 0
⇒ (x² – 4) (x² + 1) = 0
⇒ x = ± 2, ± i
Since x, y ∈ R ⇒ x = ± 2
Also 2x – y = 2y – 5 ⇒ 2x – 3y = – 5 …(1)
When x = 2
∴ from (1); 4 – 3y = – 5 ⇒ y = 3
When x = – 2
∴ from(1) ; – 4 – 3y = – 5 ⇒ y = \(\frac { 1 }{ 3 }\)

Question 6.
Write the conjugate of (6 + 5i)².
Solution:
Let z = (6 + 5i)² = 36 – 25 + 60i = 11 + 60i
∴ \(\bar { z }\) = \(\overline{11+60 i}\) = 11 – 60i

Question 7.
Write the additive inverse of the following:
(i) – 2 + 3 i
(ii) 3 – 4i
Solution:
(i) Let z = – 2 + 3i
∴ additive inverse of z = – z
= -(-2 + 3i)
= 2 – 3i

(ii) Let z = 3 – 4i
∴ additive inverse of z = – z = – (3 – 4i)
= – 3 + 4i

Question 8.
Find the multiplicative inverse of each of the following complex numbers when it exists.
(i) 2 + 2i
(ii) 0 + 0i
(iii) – 7 + 0i
(iv) – 16
(v) \(\frac{i}{1+i}\)
(vi) (1 +1)²
(vii) \(\frac{3+4 i}{4-5 i}\)
(viii) (6 + 5i)²
(ix) \(\frac{(2+3 i)(3+2 i) i}{5+i}\)
Solution:
(i) Let z = 2 + 2i
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{1}{2+2 i} \times \frac{2-2 i}{2-2 i}=\frac{2-2 i}{2^2-(2 i)^2}\)
= \(\frac{2-2 i}{4+4}=\frac{2-2 i}{8}=\frac{1}{4}-\frac{i}{4}\)

(ii) Let z = – 7 + 0i = – 7
∴Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= – \(\frac { 1 }{ 7 }\) = – \(\frac { 1 }{ 7 }\) + 0i

(iii) Let z = 0 + 0i = 0
∴ Multiplicative inverse of z does not exists, since z = 0

(iv) Let z = – 16
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{1}{-16}=\frac{-1}{16}\) + 0i

(v) Let z = \(\frac{i}{1+i}\)
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac { 1 }{ z }\)
= – i + 1 = 1 – i

(vi) Let z = (1 + i)² = 1 + i2 + 2i
= 1 – 1 + 2i = 2i
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{1}{2 i} \times \frac{i}{i}=\frac{i}{2 i^2}=\frac{i}{-2}\) = 0 – \(\frac { 1 }{ 2 }\)i

(vii) Let z = \(\frac{3+4 i}{4-5 i}\)

(viii) Let z = (6 + 5i)² = 36 – 25 + 60i = 11 + 60i

(ix) Let z = \(\frac{(2+3 i)(3+2 i) i}{5+i}\)
= \(\frac{(2 i-3)(2 i+3)}{5+i}=\frac{(2 i)^2-3^2}{5+i}\)
⇒ z = \(\frac{-4-9}{5+i}=\frac{-13}{5+i}\)
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{5+i}{-13}=\frac{-5}{13}-\frac{1}{13}\)i

Question 9.
Simplify :
(i) (1 + i)^-1
(ii) \(\sqrt{-\frac{49}{25}} \sqrt{-\frac{1}{9}}\)
(iii) \(\sqrt{-64} \cdot(3+\sqrt{-361})\)
(iv) (3 – 7i)²
(v) \(\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^2\)
(vi) \(\frac{(1-i)^3}{\left(1-i^3\right)}\)
(vii) \(\left(\frac{1+i}{1-i}\right)^{4 n+1}\) (n, is a + ve integer)
(viii) \(\frac{\sqrt{(5+12 i)}+\sqrt{(5-12 i)}}{\sqrt{(5+12 i)}-\sqrt{(5-12 i)}}\)
Solution:
(i) (1 + i)^-1 = \(\frac{1}{1+i} \times \frac{1-i}{1-i}\)
= \(\frac{1-i}{1^2-i^2}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2}\)

(ii) \(\sqrt{-\frac{49}{25}} \sqrt{-\frac{1}{9}}=\left(\frac{7}{5} i\right)\left(\frac{1}{3} i\right)\)
= \(\frac { 7 }{ 15 }\)i² = – \(\frac { 7 }{ 15 }\)

(iii) \(\sqrt{-64} \cdot(3+\sqrt{-361})\) = 8i (3 + 19i)
= 24i + 152i² = 24i – 152

(iv) (3 – 7i)² = 3² + (7i)² – 42i
= 9 – 49 – 42i = – 40 – 42i

(v) \(\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^2=\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^2\)
= \(\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2} i\right)^2+2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} i\)
= \(\frac{1}{4}-\frac{3}{4}+\frac{\sqrt{3}}{2} i=-\frac{1}{2}+\frac{\sqrt{3}}{2} i\)

(vi) \(\frac{(1-i)^3}{\left(1-i^3\right)}=\frac{1^3-i^3-3 i(1-i)}{1-i^2 \times i}\)
= \(\frac{1+i-3 i-3}{1+i}=\frac{-2-2 i}{1+i}\)
= \(\frac{-2(1+i)}{1+i}\) = – 2

(vii) \(\left(\frac{1+i}{1-i}\right)^{4 n+1}=\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^{4 n+1}\)
= \(\left[\frac{(1+i)^2}{1^2-i^2}\right]^{4 n+1}=\left[\frac{1+i^2+2 i}{1+1}\right]^{4 n+1}\)
= \(\left[\frac{1-1+2 i}{2}\right]^{4 n+1}=i^{4 n+1}\)
= i⁴ⁿ, i = (i⁴)ⁿ x i = 1ⁿ x i = i

(viii) 5 + 12i = 9 – 4+ 12i
= 3² + (2i)² + 2 x 3 x 2i
= (3 + 2i)²
and 5 – 12i = 9 – 4 – 12i
= 3² + (2i)² – 2 x 3 x 2i
= (3 – 2i)²
∴ \(\frac{\sqrt{5+12 i}+\sqrt{5-12 i}}{\sqrt{5+12 i}-\sqrt{5-12 i}}=\frac{\sqrt{(3+2 i)^2}+\sqrt{(3-2 i)^2}}{\sqrt{(3+2 i)^2}-\sqrt{(3-2 i)^2}}\)
= \(\frac{3+2 i+3-2 i}{3+2 i-(3-2 i)}\)
= \(\frac{6}{4 i}=\frac{3}{2 i} \times \frac{i}{i}=\frac{3 i}{-2}\) = – \(\frac { 3 }{ 2 }\)i

Question 10.
Prove that \(\left[\left(\frac{3+2 i}{2-5 i}\right)+\left(\frac{3-2 i}{2+5 i}\right)\right]\) is rational.
Solution:

Question 11.
Show that \(\frac{1+2 i}{3+4 i} \times \frac{1-2 i}{3-4 i}\) is real.
Solution:
\(\frac{1+2 i}{3+4 i} \times \frac{1-2 i}{3-4 i}=\frac{1^2-(2 i)^2}{3^2-(4 i)^2}\) = \(\frac{1+4}{9+16}=\frac{5}{25}=\frac{1}{5}\)
which is clearly a real number.

Question 12.
Perform the indicated operation and give your answer in the form x + yi, where x and y are real numbers and i = \(\sqrt{-1}\).
Solution:
(i) (3 + 4i)^-1
(ii) \(\frac{2-\sqrt{-25}}{1-\sqrt{-16}}\)
(iii) \(\frac{5+2 i}{-1+\sqrt{3} i}\)
(iv) \(\frac{5-3 i}{6+i}\)
(v) \((\sqrt{5}-7 i)(\sqrt{5}-7 i)^2+(-2+7 i)^2\)
Solution:

Question 13.
If x + yi = \(\frac{u+v i}{u-v i}\) prove that x² + y² = 1.
Solution:
Given x + iy = \(\frac{u+v i}{u-v i}\)
Taking modulus on both sides; we have
|x + iy| = \(\left|\frac{u+v i}{u-v i}\right|=\frac{|u+v i|}{|u-v i|}\) [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ \(\sqrt{x^2+y^2}=\frac{\sqrt{u^2+v^2}}{\sqrt{u^2+\left(-v^2\right)}}\)
⇒ \(\sqrt{x^2+y^2}\) = 1
On squaring both sides ; we have
x² + y² = 1

Question 14.
Prove that:
\([4+3 \sqrt{-20}]^{\frac{1}{2}}+[4-3 \sqrt{-20}]^{\frac{1}{2}}\) = 6.
Solution:

Question 15.
Express the following in the form a + bi :
(i) \(\sqrt{\frac{5(2+i)}{2-i}}\)
(ii) \(\frac{(3-i)^2}{2+i}\)
(iii) (1 + i)^-3
(iv) \(\left(\frac{4 i^3-i}{2 i+1}\right)^2\)
(v) \(\frac{i-1}{i+1}\)
(vi) \(\frac{2+i}{(3-i)(1+2 i)}\)
(vii) \(\frac{5}{2 i-7 i^2}\)
Solution:

Question 16.
Prove that \(\left(\frac{-1+i \sqrt{3}}{2}\right)^3\) is a positive integer.
Solution:
\(\left(\frac{-1+i \sqrt{3}}{2}\right)^3\)
= \(\frac{1}{8}\left[(-1)^3+(i \sqrt{3})^3-3 \sqrt{3} i(-1+i \sqrt{3})\right]\)
= \(\frac{1}{8}[-1-3 \sqrt{3} i+3 \sqrt{3} i+9]=\frac{8}{8}\) = 1
which is a positive integer.

Question 17.
If one of the values of x of the equation 2x² – 6x + k = 0 be \(\frac { 1 }{ 2 }\) (a + 5i), find the values of a and k.
Solution:
Given eqn. be 2x² – 6x + k = 0
⇒ x = \(\frac{6 \pm \sqrt{36-8 k}}{4}=\frac{6 \pm 2 \sqrt{9-2 k}}{4}\)
⇒ x = \(\frac{3 \pm \sqrt{9-2 k}}{2}\)
Since one of the values of x is given to be \(\frac { 1 }{ 2 }\)(a + 5i) which is only possible
if 9 – 2k = – 25 ⇒ 2k = 34 ⇒ k = 17
∴ x = \(\frac{3 \pm 5 i}{2}=\frac{1}{2}(3 \pm 5 i)\)
Hence required value of a be 3.

Question 18.
Define conjugate complex numbers and show that their sum and product are real numbers.
Solution:
Let z = x + iy be any complex number where x, y ∈ R
Then conjugate of z = \(\bar{z}=\overline{x+i y}\)
= x – iy
∴ z + \(\bar { z }\) = x + iy + x – iy = 2x, which is a real number
and z\(\bar { z }\) = (x + iy) (x – iy) = x² – (i y)²
= x² + y²
which is again a real number, since x and y are reals.

Question 19.
If \(\bar { z }\) = – z ≠ 0, show that z is necessarily a purely imaginary number.
Solution:
Let z = x + iy where x, y ∈ R, x, y are not both zero.
∴ \(\bar { z }\) = x – iy
Also, given \(\bar { z }\) = – z
⇒ x – iy = – (x + iy)
⇒ x – iy = – x – iy
⇒ 2x = 0 ⇒ x = 0
∴ z = iy which is purely an imaginary number.

Question 20.
z and z’ are complex numbers such that their product zz’ = 3 – 4i.
Given that z’ is 5 + 3i, express z in the form a + bi where a and b are rational numbers.
Solution:
Given z, z’ ∈ C
s.t zz’ = 3 – 4z …(1)
Also, z’ = 5 + 3z
∴ from (1); we have
z = \(\frac{3-4 i}{5+3 i} \times \frac{5-3 i}{5-3 i}=\frac{(3-4 i)(5-3 i)}{5^2-(3 i)^2}\)
⇒ z = \(\frac{15-9 i-20 i-12}{25+9}=\frac{3-29 i}{34}\)
⇒ z = \(\frac{3}{34}-\frac{29}{34}\) i

Question 21.
If a + bi = \(\frac{(x+i)^2}{2 x^2+1}\), prove that a² + b² = \(\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2}\).
Solution:

Question 22.
Let z₁ = 2 – i, z₂ = – 2 + i, find
(i) Re \(\left(\frac{z_1 z_2}{\bar{z}_1}\right)\),
(ii) Im \(\left(\frac{1}{z_1 \bar{z}_2}\right)\)
Solution:
Given z₁ = 2 – i, z₂ = – 2 + i
∴ \(\bar{z}_1=\overline{2-i}\) = 2 + i ;
\(\bar{z}_2=\overline{-2+i}\) = 2 + i ;

Question 23.
If z₁ = 3 + 5i and z₂ = 2 – 3i, then verify that \(\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\bar{z}_1}{\bar{z}_2}\).
Solution:

Question 24.
If x = – 2 – \(\sqrt{3}\)i, where i = \(\sqrt{-1m}\), find that value of 2x⁴ + 5x³ + 7x² – x + 41.
Solution:
Given x = – 2 – \(\sqrt{3}\)i
⇒ x + 2 = – \(\sqrt{3}\)i
On squaring both sides ; we have
(x + 2)² = (- \(\sqrt{3}\)i)² ⇒ x² + 4x + 4 = – 3
⇒ x² + 4x + 7 = 0 … (1)
Now 2x⁴ + 5x³ + 7s2 – x + 41 = 2x² (x² + 4x + 7) – 3x³ – 7x² – x + 41
= 2x² (x² + 4x + 7) – 3x (x² + 4x + 7) + 5x² – 20x + 41
= 2x² (x² + 4x + 7) – 3x (x² + 4x + 7) + 5 (x² – 4x + 7) + 6
= 2x² x 0 – 3x × 0 + 5 x 0 + 6
= 6 [using (1)]

Question 25.
If z = – 3 + \(\sqrt{-2}\), then prove that z⁴ + 5z³ + 8z² + 7z + 4 is equal to – 29.
Solution:
Given z = – 3 + \(\sqrt{-2}\) = – 3 + \(\sqrt{2}\) i
⇒ z + 3 = \(\sqrt{2}\) i; On squaring both sides ; we have
(z + 3)² = (\(\sqrt{2}\) i)² ⇒ z² + 6z + 9 = – 2
⇒ z² + 6z+ 11 =0 …(1)
∴ z⁴ + 5z³ + 8z² + 7z + 4 = z² (z² + 6z + 11) – z³ – 3z² + 7z + 4
= z²(z² + 6Z+ 11) – z (z² + 6z + 11) + 3z² + 18z + 4
= z² (z² + 6z + 11) – z (z² + 6z + 11) + 3 (z² + 6z + 11) – 29
= z² x 0 – z x 0 + 3 x 0 – 29 [using eqn. (1)]
= – 29