Question

If $(\sqrt5+\sqrt3i)^{33}=2^{49}\,z$, then modulus of the complex number $z$ is equal to

• A. $1$
• B. $\sqrt2$
• C. $2\sqrt2$
• D. $4$

Answer 1

Answer: (B)

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