Evaluate the expression ≤ft({{2}+i{2}}{2})^{64}

Correct Answer is : ei16πe^{i16π}ei16π

Correct Answer is : ei16πe^{i16π}ei16π

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UPSC NDA Math Model Paper 4

Question : 111 of 120

Marks: +1, -0

\left(\frac{\sqrt{2}+i\sqrt{2}}{2}\right)^{64}

Solution:

The given expression:

\left(\frac{\sqrt{2}+i\sqrt{2}}{2}\right)^{64}

can re-written as

=\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)^{64}

=\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)^{64}

As we know that

\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}=\cos\frac{\pi}{4}

So, we can write the given expression

\left(\frac{\sqrt{2}+i\sqrt{2}}{2}\right)^{64}

=\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{64}

As we know that,

e^{i\theta}=\cos\theta+i\sin\theta

\Rightarrow\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{64}=\left(e^{i\frac{\pi}{4}}\right)^{64}

\Rightarrow\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{64}=e^{i16\pi}

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