Find, (i) The square root of 7 + 24 i (ii) i + - i

Let 7 + 24 i = x + i y ⇒ − 7 + 24 i = x 2 − y 2 + 2 i x y Equating the real and imaginary part we get, x 2 − y 2 = − 7 & 2 x y = 24 We know, x 2 + y 2 = x 2 − y 2 + 4 x 2 y 2 = ( − 7 ) 2 + ( 24 ) 2 = 25 Solving x 2 − y 2 = − 7 & x 2 + y 2 = 25 we get, x = ± 3 & y = ± 4 As x y positive, x & y must have same sign So, 7 + 24 i = ± 4 + i 3 . i i i + − i = e i π 2 + − e i π 2 = ± e i π 4 + ± i e i π 4 = ± e i π 4 ( 1 + i ) = ± e i π 4 2 e i π 4 = ± 2 e i π 2 = ± i 2 Again, Assume, i + - i = x ⇒ i + - i 2 = x 2 ⇒ i - i + 2 i - i = x 2 ⇒ 2 1 = x 2 ⇒ x = ± 2 .

Find, (i) The square root of 7 + 24 i (ii) i + - i

Let 7 + 24 i = x + i y ⇒ − 7 + 24 i = x 2 − y 2 + 2 i x y Equating the real and imaginary part we get, x 2 − y 2 = − 7 & 2 x y = 24 We know, x 2 + y 2 = x 2 − y 2 + 4 x 2 y 2 = ( − 7 ) 2 + ( 24 ) 2 = 25 Solving x 2 − y 2 = − 7 & x 2 + y 2 = 25 we get, x = ± 3 & y = ± 4 As x y positive, x & y must have same sign So, 7 + 24 i = ± 4 + i 3 . i i i + − i = e i π 2 + − e i π 2 = ± e i π 4 + ± i e i π 4 = ± e i π 4 ( 1 + i ) = ± e i π 4 2 e i π 4 = ± 2 e i π 2 = ± i 2 Again, Assume, i + - i = x ⇒ i + - i 2 = x 2 ⇒ i - i + 2 i - i = x 2 ⇒ 2 1 = x 2 ⇒ x = ± 2 .

Simplify and express the result in the form of a + b i : (i) - i ( 9 + 6 i ) ( 2 - i ) - 1 (ii) 4 i 3 - i 2 i + 1 2 (iii) 1 ( 1 - cosθ ) + 2 i sinθ (iv) ( 3 + i ) e - i π 6

i − i ( 9 + 6 i ) 2 − i = − 9 i − 6 i 2 2 − i = 6 − 9 i 2 − i = 6 − 9 i 2 − i × 2 + i 2 + i = ( 6 − 9 i ) ( 2 + i ) ( 2 ) 2 − ( i ) 2 = 12 + 6 i − 18 i − 9 i 2 4 + 1 = 12 + 9 − 12 i 5 = 21 5 - 12 5 i ii 4 i 3 - i 2 i + 1 2 = 4 i 3 - i 1 - 2 i 2 i + 1 1 - 2 i 2 = - 5 i 1 - 2 i 1 + 4 2 = - 5 i - 2 i 2 5 2 = - 5 i + 2 5 2 = 2 + i 2 = 4 - 1 + 4 i = 3 + 4 i iii 1 ( 1 − cos θ ) + 2 i sin θ = 1 2 sin 2 θ 2 + 4 i sin θ 2 cos θ 2 = 1 2 sin θ 2 sin θ 2 + 2 i cos θ 2 × sin θ 2 − 2 i cos θ 2 sin θ 2 − 2 i cos θ 2 = sin θ 2 − 2 i cos θ 2 2 sin θ 2 sin 2 θ 2 + 4 cos 2 θ 2 = sin θ 2 − 2 i cos θ 2 2 sin θ 2 1 + 3 cos 2 θ 2 = 1 2 1 + 3 cos 2 θ 2 + i - cot θ 2 1 + 3 cos 2 θ 2 iv ( 3 + i ) e - i π 6 = ( 3 + i ) cos π 6 - isin π 6 = ( 3 + i ) 3 2 - i 2 = ( 3 ) 3 2 - i 2 + i 3 2 - i 2 = 3 2 - i 3 2 + i 3 2 + 1 2 = 2 .

Find the modulus, argument, and the principal argument of the complex numbers. (i) z = 1 + cos 18 π 25 + i sin 18 π 25 (ii) z = - 2 cos 30 ° + i sin 30 ° (iii) ( tan 1 - i ) 2 (iv) i - 1 i 1 - cos 2 π 5 + sin 2 π 5

i z = 1 + cos 18 π 25 + i sin 18 π 25 ⇒ z = 2 cos 2 9 π 25 + 2 isin 9 π 25 cos 9 π 25 = 2 cos 9 π 25 cos 9 π 25 + isin 9 π 25 So, the modulus of the function = 2 cos 9 π 25 The principal argument is 9 π 25 and the principal argument is 9 π 25 + 2 k π , k ∈ I ii z = - 2 cos 30 ° + i sin 30 ° ⇒ z = 2 - 3 2 - i 2 ⇒ z = 2 cos 5 π 6 + i sin 5 π 6 So, the modulus of the function = 2 The principal argument is 5 π 6 and the argument is 5 π 6 + 2 k π , k ∈ I iii tan 1 - i 2 = tan 2 1 - 1 - 2 i tan 1 So, the modulus of the function is tan 2 1 - 1 2 - 4 tan 2 1 = sec 2 1 The argument of the function is tan - 1 2 tan 1 1 - tan 2 1 = tan - 1 tan 2 = 2 - π and the argument is 2 k π + 2 - π , k ∈ I iv z = i − 1 i 1 − cos 2 π 5 + sin 2 π 5 = i − 1 i 2 sin 2 π 5 + 2 sin π 5 cos π 5 = i − 1 2 sin π 5 cos π 5 + isin π 5 | z | = | i − 1 | 2 s i n π 5 c o s π 5 + i s i n π 5 = 2 2 sin π 5 cos π 5 + isin π 5 = 1 2 cosec π 5 argz = arg i − 1 2 sin π 5 cos π 5 + isin π 5 = arg ( − 1 + i ) − arg 2 sin π 5 − arg cos π 5 + isin π 5 = 3 π 4 − 0 − π 5 = 11 π 20 So, the argument of function is 11 π 20 + 2 k π , k ∈ I Hence, arg z = 2 kπ + 11 π 20 , principal arg = 11 π 20

If ( 3 + i ) 100 = 2 99 ( a + i b ) , then find (i) a 2 + b 2 (ii) b

We know, the cube root of unity ω = - 1 + i 3 2 So, ( 3 + i ) = 2 ω i = - 2 ω i Thus, ( 3 + i ) 100 = 2 100 × ω 100 × i 100 = 2 100 × ω = 2 100 × - 1 + i 3 2 = 2 99 × - 1 + i 3 Comparing with a & b we get, a = - 1 & b = 3 So, a 2 + b 2 = 1 + 3 = 4

Show that e i 2 m θ i cotθ + 1 i cotθ - 1 m = 1 .

e 2 mi θ icot θ + 1 icot θ − 1 m = e 2 mi θ i ( cot θ − i i ( cot θ + i m = e 2 mi θ cot θ − i cot θ + i m = e 2 mi θ cos θ − isin θ cos θ + isin θ m = e 2 mi θ e − i θ e i θ m = e 2 mi θ ( e − 2 iθ ) m = e 2 mi θ e − 2 mi θ = e 0 = 1

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 5

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Find,
(i) The square root of $7 + 24 i$

(ii) $\sqrt{i} + \sqrt{- i}$

Important Questions on Complex Numbers

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 6

Solve the following for

z

$z^{2} - (3 - 2 i) z = (5 i - 5)$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 7

Simplify and express the result in the form of

a + b i :

(i) $- i (9 + 6 i) (2 - i)^{- 1}$
(ii) ${(\frac{4 i^{3} - i}{2 i + 1})}^{2}$
(iii) $\frac{1}{(1 - cosθ) + 2 i sinθ}$
(iv) $(\sqrt{3} + i) e^{- i \frac{π}{6}}$