Question

If $z_1=\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}+i\sin \frac{\mathrm{\pi }}{4}\right)$, $z_2=\sqrt{3}\left(\cos \frac{\mathrm{\pi }}{3}+i\sin \frac{\mathrm{\pi }}{3}\right)$, then $\left|z_1{z}_2\right|$ is equal to

• A. $6$
• B. $\sqrt{2}$
• C. $\sqrt{6}$
• D. $\sqrt{3}$
• E. $\sqrt{2}+\sqrt{3}$

Answer 1

The correct option is C

$\sqrt{6}$

Explanation for the correct option.

Find the value of $\left|z_1{z}_2\right|$.

The magnitude of $z_1=\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}+i\sin \frac{\mathrm{\pi }}{4}\right)$is given as:

$\begin{array}{rcl}\left|z_1\right|& =& \sqrt{{\left(\sqrt{2}\right)}^2\left({\cos }^2\frac{\mathrm{\pi }}{4}+{\sin }^2\frac{\mathrm{\pi }}{4}\right)}\\ & =& \sqrt{2\times 1}\left[{\cos }^{2}A+{\sin }^{2}A=1\right]\\ & =& \sqrt{2}\end{array}$

The magnitude of $z_2=\sqrt{3}\left(\cos \frac{\mathrm{\pi }}{3}+i\sin \frac{\mathrm{\pi }}{3}\right)$ is given as:

$\begin{array}{rcl}\left|z_2\right|& =& \sqrt{{\left(\sqrt{3}\right)}^2\left({\cos }^2\frac{\mathrm{\pi }}{3}+{\sin }^2\frac{\mathrm{\pi }}{3}\right)}\\ & =& \sqrt{3\times 1}\left[{\cos }^{2}A+{\sin }^{2}A=1\right]\\ & =& \sqrt{3}\end{array}$

So the value of the expression $\left|z_1{z}_2\right|$ is given as:

$\begin{array}{rcl}\left|z_1{z}_2\right|& =& \left|z_1\right|·\left|z_2\right|\left[\left|ab\right|=\left|a\right|·\left|b\right|\right]\\ & =& \sqrt{2}·\sqrt{3}\left[\left|z_1\right|=\sqrt{2},\left|z_2\right|=\sqrt{3}\right]\\ & =& \sqrt{2\times 3}\left[\sqrt{a}·\sqrt{b}=\sqrt{ab}\right]\\ & =& \sqrt{6}\end{array}$

So the value of $\left|z_1{z}_2\right|$ is $\sqrt{6}$.

Hence, the correct option is C.