Test Index

Complex Numbers

Section: Mathematics

© examsnet.com

Question : 4 of 50

Marks: +1, -0

(1+\sqrt{5}+i\sqrt{10-2\sqrt{5}})^5=

[AP EAPCET 24 May 2025 S1]

$\; 1024$
$\; -1024$
$\; 512$
$\; -512$

Solution:

Let

z=1+\sqrt{5}+i\sqrt{10-2\sqrt{5}}

\text{ So, } |z| \; = \sqrt{(1+\sqrt{5})^2+(\sqrt{10-2\sqrt{5}})^2}

\; = \sqrt{1+5+2\sqrt{5}+10-2\sqrt{5}}

\; = \sqrt{16+2\sqrt{5}-2\sqrt{5}} = \sqrt{16} = 4

\; \text{ And } \; \theta \; = \arg(z) = \tan^{-1} \left( \frac{\sqrt{10-2\sqrt{5}}}{1+\sqrt{5}} \right)

\; = \tan^{-1} \left( \tan \left( \frac{\pi}{5} \right) \right) = \frac{\pi}{5}

So,

z = |z| \cdot (\cos \theta + i \sin \theta)

\; = 4 \left[ \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right]

\therefore z^5 \; = 4^5 \left[ \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right]^5

\; = 1024 \left[ \cos \left( 5 \cdot \frac{\pi}{5} \right) + i \sin \left( 5 \cdot \frac{\pi}{5} \right) \right]

\; = 1024[ \cos \pi + i \sin \pi ]

\; = 1024[-1+0] = -1024

© examsnet.com

Go to Question:

Prev Question

Next Question