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Complex Numbers

Section: Mathematics

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Question : 17 of 50

Marks: +1, -0

\left(\sqrt{\sqrt{2}+1}+i\sqrt{\sqrt{2}-1}\right)^8=

[AP EAPCET 21 May 2025 S1]

64
64 i
-64
$-64i$

Solution:

Let,

Z=\sqrt{\sqrt{2}+1}+i\sqrt{\sqrt{2}-1}

\; \Rightarrow \;\; |Z|=\sqrt{\sqrt{2}+1+\sqrt{2}-1}=\sqrt{2\sqrt{2}}

\; \;\; \cos \theta=\frac{\sqrt{\sqrt{2}+1}}{\sqrt{2\sqrt{2}}}, \sin \theta=\frac{\sqrt{\sqrt{2}-1}}{\sqrt{2\sqrt{2}}}

\; \therefore \;\; \cos^2 \theta=\frac{\sqrt{2}+1}{2\sqrt{2}}, \sin^2 \theta=\frac{\sqrt{2}-1}{2\sqrt{2}}

\; \Rightarrow \;\frac{1+\cos 2\theta}{2}=\frac{\sqrt{2}+1}{2\sqrt{2}}

\; \Rightarrow \cos 2\theta=\frac{1}{\sqrt{2}}

\Rightarrow 2\theta=\frac{\pi}{4}

(since, real and imaginary part are positive.)

Now,

\; Z^8=\left(\sqrt{2\sqrt{2}}\right)^8 \left(\cos\left(8 \times \; \frac{\pi}{8}\right)+i\sin\left(8 \times \; \frac{\pi}{8}\right)\right)

\; =\left(2\sqrt{2}\right)^4\left(\cos\pi+i\sin\pi\right)

\; = \left(2^{\frac{3}{2}} \times 4\right)(-1+0)

\; =2^6(-1)=-64

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