Question

If $z=\frac{{\left(\sqrt{3}+i\right)}^3{\left(3i+4\right)}^2}{{\left(8+6i\right)}^2}$, then $\left|z\right|$ is equal to

Answer 1

Find the modulus of $z$.

For the complex number $z=\frac{{\left(\sqrt{3}+i\right)}^3{\left(3i+4\right)}^2}{{\left(8+6i\right)}^2}$, the modulus is given as

$\begin{array}{rcl}\left|z\right|& =& \left|\frac{{\left(\sqrt{3}+i\right)}^3{\left(3i+4\right)}^2}{{\left(8+6i\right)}^2}\right|\\ & =& \frac{\left|{\left(\sqrt{3}+i\right)}^3\right|\left|{\left(3i+4\right)}^2\right|}{\left|{\left(8+6i\right)}^2\right|}\left[\left|\frac{a}{b}\right|=\frac{\left|a\right|}{\left|b\right|},\left|ab\right|=\left|a\right|\left|b\right|\right]\\ & =& \frac{{\left|\sqrt{3}+i\right|}^3{\left|3i+4\right|}^2}{{\left|8+6i\right|}^2}\left[\left|a^n\right|={\left|a\right|}^n\right]\\ & =& \frac{{\left(\sqrt{{\left(\sqrt{3}\right)}^2+1^2}\right)}^3{\left(\sqrt{3^2+4^2}\right)}^2}{{\left(\sqrt{8^2+6^2}\right)}^2}\left[\left|a+ib\right|=\sqrt{a^2+b^2}\right]\\ & =& \frac{{\left(\sqrt{4}\right)}^3{\left(\sqrt{25}\right)}^2}{{\left(\sqrt{100}\right)}^2}\\ & =& \frac{2^3\times 25}{100}\\ & =& 2\end{array}$

The value of the modulus of complex number $z=\frac{{\left(\sqrt{3}+i\right)}^3{\left(3i+4\right)}^2}{{\left(8+6i\right)}^2}$ is $2$.

Hence, the value of $\left|z\right|$ is $2$.