Question

Convert the given complex number into polar form: $1-i$

Answer 1

Given: $1-i$

Let $z=1-i$

In polar form, complex numbers can be expressed as $z=\left(r\cos \theta \right)+i\left(r\sin \theta \right)$.

Comparing the given complex number with the standard form(polar form) we will get

$$\begin{gathered} r\cos \theta =1....\left(I\right) \\ r\sin \theta =(-1)....\left(II\right) \end{gathered}$$

STEP 1: Find out the value of r as follows

On squaring and adding, we obtain:

$$\begin{gathered} r^2{\cos }^2\theta ={\left(1\right)}^2 \\ {r}^2{\sin }^2\theta ={(-1)}^2 \\ {r}^2{\cos }^2\theta +r^2{\sin }^2\theta ={\left(1\right)}^2+{(-1)}^2 \\ {r}^2({\cos }^2\theta +{\sin }^2\theta )=2(Since,{\cos }^2\theta +{\sin }^2\theta =1) \\ \Rightarrow r^2=2 \\ \Rightarrow r=\sqrt{2}(Since,r>0) \end{gathered}$$

Substituting the value of $r$ in (I) & (II) we will get

$$\begin{gathered} \sqrt{2}\cos \theta =1or\cos \theta =\frac{1}{\sqrt{2}}....\left(III\right) \\ \sqrt{2}\sin \theta =-1or\sin \theta =-\frac{1}{\sqrt{2}}....\left(IV\right) \end{gathered}$$

STEP 2: Find out the value of $\theta$

From (III) & (IV) we can conclude that

$\theta$ lies in the IV^th quadrant, $\theta =-\frac{\mathrm{\pi }}{4}$

The polar form for the given complex number will be

$$\begin{gathered} z=1-i \\ =\left(r\cos \theta \right)+i\left(r\sin \theta \right) \\ =\sqrt{2}\cos \left(-\frac{\mathrm{\pi }}{4}\right)+\sqrt{2}i\sin \left(-\frac{\mathrm{\pi }}{4}\right) \\ =\sqrt{2}\left\{\cos \left(-\frac{\mathrm{\pi }}{4}\right)+i\sin \left(-\frac{\mathrm{\pi }}{4}\right)\right\} \end{gathered}$$

Therefore in polar form $1-i=\sqrt{2}\left\{\cos \left(-\frac{\mathrm{\pi }}{4}\right)+i\sin \left(-\frac{\mathrm{\pi }}{4}\right)\right\}$