Question

If ${{(\sqrt{5}+\sqrt{3}i)}^{33}}={{2}^{49}}z,$ then modulus of the complex number $z$ is equal to

• A. 1
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. 4

Answer 1

Answer: (B)

$\sqrt{2}$

Answer 2

Given, ${{(\sqrt{5}+\sqrt{3}i)}^{33}}={{2}^{49}}z$ Let $\sqrt{5}=r\cos \theta ,\sqrt{3}=r\sin \theta$
$\therefore$ ${{r}^{2}}=5+3$
$\Rightarrow$ $r=2\sqrt{2}$
$\therefore$ ${{(r\cos \theta +ir\sin \theta )}^{33}}={{2}^{49}}z$
$\Rightarrow$ $|{{r}^{33}}(cos33\theta +i\sin 33\theta )=|{{2}^{49}}z|$
$\Rightarrow$ ${{(2\sqrt{2})}^{33}}|\cos 33\theta +i\sin 33\theta |={{2}^{49}}|z|$
$\Rightarrow$ ${{2}^{\frac{99}{2}}}(1)={{2}^{49}}|z|$
$\Rightarrow$ $|z|={{2}^{\frac{99}{2}-49}}$
$\Rightarrow$ $|z|=\sqrt{2}$