Question

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

• A. $|z|=\frac{1}{\sqrt{2}};amp\left(z\right)=-\frac{3\pi }{4}$
• B. $|z|=\frac{1}{2};amp\left(z\right)=\frac{3\pi }{4}$
• C. $|z|=\frac{1}{2};amp\left(z\right)=-\frac{3\pi }{4}$
• D. $|z|=\frac{1}{\sqrt{2}};amp\left(z\right)=\frac{3\pi }{4}$

Answer 1

The correct option is D $|z|=\frac{1}{\sqrt{2}};amp\left(z\right)=\frac{3\pi }{4}$

Let $z=\frac{1+2i}{1-3i},$ then

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

$z=\frac{1+3i+2i+6i^2}{(1{)}^2+(3{)}^2}$

$z=\frac{1+5i+6(-1)}{1+9}$

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

$z=\frac{-1}{2}+\frac{1}{2}i$

Let $z=r\cos \theta +ir\sin \theta$

$r\cos \theta =\frac{-1}{2}$ and $r\sin \theta =\frac{1}{2}$

On squaring and adding, we obtain

$r^2({\cos }^2\theta +{\sin }^2\theta )={\left(\frac{-1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$

$\Rightarrow$ $r^2=\frac{1}{4}+\frac{1}{4}$

$\Rightarrow$ $r^2=\frac{1}{2}$

$\therefore$ $r=\frac{1}{\sqrt{2}}$ [ Conventionally, $r>0$ ]

$\therefore$ $\frac{1}{\sqrt{2}}\cos \theta =\frac{-1}{2}$ and $\frac{1}{\sqrt{2}}\sin \theta =\frac{1}{2}$

$\Rightarrow$ $\cos \theta =\frac{-1}{\sqrt{2}}$ and $\sin \theta =\frac{1}{\sqrt{2}}$

$\therefore$ $\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$ [ As $\theta$ lies in the $II$ quadrant ]

$\therefore$ The modulus and amplitude of the given complex number are $\frac{1}{\sqrt{2}}$ and $\frac{3\pi }{4}$