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)

Given that, ${\left(\sqrt{5}+\sqrt{3}i\right)}^{33}=2^{49}z$
Let $\sqrt{5}=rcos\theta ,\sqrt{3}=rsin\theta$
$\therefore r^2=5+3$
$\Rightarrow r=2\sqrt{2}$
$\therefore {\left(rcos\theta +irsin\theta \right)}^{33}=2^{49}z$
$\Rightarrow \left|r^{33}\left(cos33\theta +isin33\theta \right)\right|=\left|2^{49}z\right|$
$\Rightarrow {\left(2\sqrt{2}\right)}^{33}\left|cos33\theta +isin33\theta \right|=2^{49}\left|z\right|$
${\left(2^{3/2}\right)}^{33}=2^{49}\left|z\right|$
$\Rightarrow \left|z\right|=2^{\frac{99}{2}-49}=2^{\frac{1}{2}}=\sqrt{2}$