The smallest positive integral value of n such that [(1+ (π)/(8)+i (π)/(8))/(1+ (π)/(8)-i (π)/(8))]^n is purely imaginary, is equal to

Correct answer: A. 4 [(1+ (π)/(8)+i (π)/(8))/(1+ (π)/(8)-i (π)/(8))]^n=[(1+ α+i α)/(1+ α-i α)]^n[ α=(π)/(2)-(π)/(8)]=[(2 ^2 (α)/(2)+2 i (α)/(2) (α)/(2))/(2 ^2 (α)/(2)-2 i (α)/(2) (α)/(2))]^n=[( (α)/(2)+i (α)/(2))/( (α)/(2)-i (α)/(2))]_ - ^n=(e^ 2 i (α)/(2) ]^n=e^ i n α =e^ i n((3 π)/(8)) = (3 n π)/(8)+i (3 n π)/(8) For n=4 we get imaginary part.

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Complex Number

BITSAT 2022 · Mathematics · Complex Number

The smallest positive integral value of n such that [(1+ (π)/(8)+i...

The smallest positive integral value of

n

such that

\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^n

is purely imaginary, is equal to

Options

A $4$
B $3$
C $2$
D $8$

Correct answer

A. $4$

Step-by-step solution

\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^n=\left[\frac{1+\cos \alpha+i \sin \alpha}{1+\cos \alpha-i \sin \alpha}\right]^n\left[\because \alpha=\frac{\pi}{2}-\frac{\pi}{8}\right]=\left[\frac{2 \cos ^2 \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^2 \frac{\alpha}{2}-2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right]^n=\left[\frac{\cos \frac{\alpha}{2}+i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-i \sin \frac{\alpha}{2}}\right]_{-}^n=\left(e^{2 i \frac{\alpha}{2}}\right]^n=e^{i n \alpha}=e^{i n\left(\frac{3 \pi}{8}\right)}=\cos \frac{3 n \pi}{8}+i \sin \frac{3 n \pi}{8}

For

n=4

we get imaginary part.

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