Find the modulus and amplitude of $\frac{1+3 \sqrt{3} \mathrm{i}}{\sqrt{3}+2 \mathrm{i}}$.

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written 3.0 years ago by

prajapatijaimin • 3.3k

• modified 3.0 years ago

Solution:

$Let, \mathrm{z}=\frac{\overline{1}+3 \sqrt{3} \mathrm{i}}{\sqrt{3}+2 \mathrm{i}}\\$

$\begin{aligned} =& \frac{1+3 \sqrt{3} \mathrm{i}}{\sqrt{3}+2 \mathrm{i}} \times \frac{\sqrt{3}-2 \mathrm{i}}{\sqrt{3}-2 \mathrm{i}}\\ =& \frac{\sqrt{3}-2 \mathrm{i}+9 \mathrm{i}-6 \sqrt{3} \mathrm{i}^{2}}{(\sqrt{3})^{2}-(2 \mathrm{i})^{2}} \\ =& \frac{\sqrt{3}+7 \mathrm{i}+6 \sqrt{3}}{3+4} \\ =& \frac{7 \sqrt{3}+7 \mathrm{i}}{7} \\ =& \frac{7(\sqrt{3}+\mathrm{i})}{7} \\ \mathrm{z}=& \sqrt{3}+\mathrm{i}=\mathrm{a}+\mathrm{ib} \text { form } \end{aligned}$

$Here, \mathrm{a}=\sqrt{3} \& \mathrm{~b}=1\\$

Modulus:

$:|z|=\sqrt{a^{2}+b^{2}}=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{3+\sqrt{1}}=4=2$

Amplitude:

$\theta=\tan ^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$

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