Question

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

• A. $1$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $8$

Answer 1

The correct option is B

$\sqrt{2}$

Explanation for the correct option:

Step : Using the property of modulus of the complex number $z$, that is $\left|z\right|$

The property complex number if $z=x+iy$ then $\left|z\right|=\sqrt{x^2+y^2}$ and $\left|z^n\right|={\left|z\right|}^n$

Using the above property, solve the given ${(\sqrt{5}+\sqrt{3}i)}^{33}=2^{49}z$

Take modulus on both the sides of the above relation

$\begin{array}{l}\Rightarrow \left|{(\sqrt{5}+\sqrt{3}i)}^{33}\right|=\left|2^{49}z\right|\\ \Rightarrow \left|{(\sqrt{5}+\sqrt{3}i)}^{33}\right|=2^{49}\left|z\right|\\ {\Rightarrow \left(\sqrt{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2}\right)}^{33}=2^{49}\left|z\right|\\ {\Rightarrow \left(\sqrt{5+3}\right)}^{33}=2^{49}\left|z\right|\\ {\Rightarrow \left(\sqrt{8}\right)}^{33}=2^{49}\left|z\right|\\ \Rightarrow 8^{\frac{33}{2}}=2^{49}\left|z\right|\end{array}$

$\begin{array}{l}\Rightarrow {\left(2^3\right)}^{\frac{33}{2}}=2^{49}\left|z\right|\\ \Rightarrow 2^{\frac{99}{2}}=2^{49}\left|z\right|\\ \Rightarrow \left|z\right|=2^{(\frac{99}{2}-49)}\\ \Rightarrow \left|z\right|=2^{\left(\frac{99-98}{2}\right)}\\ \Rightarrow \left|z\right|=2^{\frac{1}{2}}\\ \Rightarrow \left|z\right|=\sqrt{2}\end{array}$

Hence, the correct option is (B).