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
• E. 8

Answer 1

Answer: (B)

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