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

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

Express each of the following in the form b or bi, where b is a real number:

Question 1.
3i . 2
Solution:
3i . 2 = 6i

Question 2.
i (- i)
Solution:
i (- i) = – i² = – (- 1) = 1 + 0i

Question 3.
– i (- i)
Solution:
– i (- i) = i² = – 1

Question 4.
5i (- 8i)
Solution:
5i (- 8i) = – 40i² = – 40 (- 1) = 40

Question 5.
\(\frac { 20i }{ 4 }\)
Solution:
\(\frac { 20i }{ 4 }\) = 5i

Question 6.
\(\sqrt{-25}\)
Solution:
\(\sqrt{-25}=\sqrt{5 \times 5} \sqrt{-1}\) = 5i

Question 7.
\(\sqrt{-8}\)
Solution:
\(\sqrt{-8}=\sqrt{2 \times 2} \sqrt{-2}=2 \sqrt{2} i\)

Question 8.
\(\sqrt{\frac{-1}{3}}\)
Solution:
\(\sqrt{\frac{-1}{3}}=\sqrt{\frac{1}{3}} i=\sqrt{\frac{1}{3} \times \frac{3}{3}} i=\frac{\sqrt{3}}{3} i\)

Question 9.
\(\frac{1}{2} \sqrt{\frac{-3}{4}}\)
Solution:
\(\frac{1}{2} \sqrt{\frac{-3}{4}}=\frac{1}{2} \times \frac{1}{2} \sqrt{3} i=\frac{\sqrt{3}}{4} i\)

Question 10.
\(\frac { 6 }{ -i }\)
Solution:
\(\frac{6}{-i}=\frac{6}{-i} \times \frac{i}{i}=\frac{6 i}{-i^2}=\frac{6 i}{-(-1)}\) = 6i

Question 11.
\(\sqrt{-144}\)
Solution:
\(\sqrt{-144}=\sqrt{12 \times 12} \sqrt{-1}\) = 12i

Question 12.
\(\frac { x }{ i }\)
Solution:
\(\frac{x}{i}=\frac{x}{i} \times \frac{i}{i}=\frac{i x}{i^2}\) = – i x

Question 13.
i¹³
Solution:
i¹³ = i¹² i = (i⁴)³ i = 1³ i = i [∵ i⁴ = 1]

Question 14.
i²⁸
Solution:
i²⁸ = (i⁴)⁷ = 1 [∵ i⁴ = (i²)² = (- 1)² = 1]

Question 15.
i¹⁸
Solution:
i¹⁸ = (i⁴)⁴ i² = 1⁴ x (- 1) = – 1

Question 16.
i²³
Solution:
i²³ = (i⁴)⁵ i³ = 1⁵ x i² x i = 1 x (- 1) x i = – i

Question 17.
\(\sqrt{-4}+\sqrt{-16}-\sqrt{-25}\)
Solution:
\(\sqrt{-4}+\sqrt{-16}-\sqrt{-25}\) = 2i + 4i – 5i
= 6i – 5i = i

Question 18.
\(\sqrt{-20}+\sqrt{-12}\)
Solution:
\(\sqrt{-20}+\sqrt{-12}=2 \sqrt{5} i+2 \sqrt{3} i\)
= 2(\(\sqrt{5}+\sqrt{2}\))i

Question 19.
– \(\sqrt{\frac{-7}{4}}-\sqrt{\frac{-1}{7}}\)
Solution:

Question 20.
\(\frac{\sqrt{-2}}{\sqrt{-8}}\)
Solution:
\(\frac{\sqrt{-2}}{\sqrt{-8}}=\frac{\sqrt{2} i}{2 \sqrt{2} i}=\frac{1}{2}\)

Question 21.
\(\frac{1}{i}+\frac{1}{i^2}+\frac{1}{i^3}+\frac{1}{i^4}\)
Solution:
\(\frac{1}{i}+\frac{1}{i^2}+\frac{1}{i^3}+\frac{1}{i^4}=\frac{1}{i}+\frac{1}{-1}+\frac{1}{-i}+1\)
= \(\frac{1}{i}-1-\frac{1}{i}+1\) = 0

Question 22.
\(\frac{1}{i}-\frac{1}{i^2}+\frac{1}{i^3}-\frac{1}{i^4}\)
Solution:
\(\frac{1}{i}-\frac{1}{i^2}+\frac{1}{i^3}-\frac{1}{i^4}=\frac{1}{i}+1-\frac{1}{i}-1\) = 0 [∵ i² = – 1 and i⁴ = 1]

Question 23.
i + 2i² + 3i³ + i⁴
Solution:
i + 2i² + 3i³ + i⁴ = i + 2 (- 1) + 3i (- 1) + 1 = i – 2 – 3i + 1 = – 2i – 1

Question 24.
\(\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3\)
Solution:

Question 25.
\(\sqrt{\frac{-x}{4}}+\sqrt{\frac{-x}{16}}-\sqrt{\frac{-x}{64}}\),
where x is a positive real number.
Solution:

Question 26.
\(\sqrt{-5 x^8}-\sqrt{-20 x^8}+\sqrt{-45 x^8}\), where x a positive real number.
Solution:
\(\sqrt{-5 x^8}-\sqrt{-20 x^8}+\sqrt{-45 x^8}\)
= \(\sqrt{5} x^4 i-2 \sqrt{5} x^4 i+3 \sqrt{5} x^4 i\)
= \(2 \sqrt{5} i x^4\)

Question 27.
If i = \(\sqrt{-1}\) prove the following :
(x + 1 + i) (x + 1 – i) (x – 1 – i) = x⁴ + 4.
Solution:
L.H.S = (x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i)
= [(x + 1)² – i²] [(x – 1)² – i²]
= [x² + 2x + 1 + 1] [x² – 2x + 1 + 1]
= (x² + 2x + 2) (x² – 2x + 2)
= (x² + 2 + 2x) (x² + 2 – 2x)
= (x² + 2)² – (2x)²
= x⁴ + 4x² + 4 – 4x²
= x⁴ + 4 = R.H.S