Question

Find the modulus and argument of the complex number ${\displaystyle \frac{1+2\text{i}}{1-3\text{i}}}$

Answer 1

Let z = ${\displaystyle \frac{1+2\text{i}}{1-3\text{i}}}$

= ${\displaystyle \frac{1+2\text{i}}{1-3\text{i}}\times \frac{1+3\text{i}}{1+3\text{i}}}$

= ${\displaystyle \frac{1+3\text{i}+2\text{i}+6{\text{i}}^2}{1-9{\text{i}}^2}}$

= ${\displaystyle \frac{1+5\text{i}-6}{1+9}}$   ...[∵ i² = – 1]

∴ z = ${\displaystyle \frac{-5+5\text{i}}{10}}$

= ${\displaystyle -\frac{1}{2}+\frac{1}{2}\text{i}}$

This is of the form a + bi, where a = ${\displaystyle -\frac{1}{2}}$, b = ${\displaystyle \frac{1}{2}}$

∴ modulus = r 

= ${\displaystyle \sqrt{{\text{a}}^2+{\text{b}}^2}}$

= ${\displaystyle \sqrt{{(-\frac{1}{2})}^2+{\left(\frac{1}{2}\right)}^2}}$

= ${\displaystyle \sqrt{\frac{1}{4}+\frac{1}{4}}}$

= ${\displaystyle \frac{1}{\sqrt{2}}}$

If θ is the argument, then

cos θ = ${\displaystyle \frac{\text{a}}{\text{r}}}$

= ${\displaystyle \frac{(-\frac{1}{2})}{\left(\frac{1}{\sqrt{2}}\right)}}$

= ${\displaystyle -\frac{1}{\sqrt{2}}}$

and sin θ = ${\displaystyle \frac{\text{b}}{\text{r}}}$

= ${\displaystyle \frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{\sqrt{2}}\right)}}$

= ${\displaystyle \frac{1}{\sqrt{2}}}$

${\displaystyle \therefore \theta =\frac{3\pi }{4} ...\left[\begin{array}{c}\because \cos \frac{3\pi }{4}=\cos (\pi -\frac{\pi }{4})=-\cos \frac{\pi }{4}=\\ -\frac{1}{\sqrt{2}}\text{and}\sin \frac{3\pi }{4}=\sin (\pi -\frac{\pi }{4})=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\end{array}\right]}$

Hence, modulus = ${\displaystyle \frac{1}{\sqrt{2}}}$ and argurement = ${\displaystyle \frac{3\pi }{4}}$