Question

Write $\left( 1-i \right)$ in polar form.

Answer 1

Hint: For solving this problem, we first express the general polar form of a complex number. Now, by comparing our given number with general form, we formulate two equations to evaluate the argument and modulus. By using this methodology, we can easily solve our problem.

Complete step-by-step solution -

According to the problem statement, we are given a complex number in rectangular form as: $\left( 1-i \right)\ldots (1)$
The expression of a complex number in polar form can be expanded as:
$Z=r\cos \theta +ir\sin \theta \ldots (2)$
Now, comparing equation (1) and equation (2), we can relate the variables as
$\begin{align}
  & r\cos \theta =1\ldots (3) \\
 & r\sin \theta =-1\ldots (4) \\
\end{align}$
Dividing equation (3) and equation (4), we get
$\dfrac{r\sin \theta }{r\cos \theta }=\dfrac{-1}{1}$
Cancelling r from numerator and denominator and rewriting $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $, we get
$\begin{align}
  & \tan \theta =-1 \\
 & \therefore \theta ={{\tan }^{-1}}(-1) \\
\end{align}$
Now, by using the property of inverse of tan function ${{\tan }^{-1}}(-x)=-{{\tan }^{-1}}(x)\text{ and ta}{{\text{n}}^{-1}}(1)=\dfrac{\pi }{4}$, we get
$\theta =-\dfrac{\pi }{4}$
Putting the obtained value of argument in equation (3), we get
$r\cos \left( \dfrac{-\pi }{4} \right)=1$
By using the conversion of cosine function $\cos \left( \dfrac{-\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ , we get
$\begin{align}
  & r\times \dfrac{1}{\sqrt{2}}=1 \\
 & r=\sqrt{2} \\
\end{align}$
So, the polar form of is $Z=\sqrt{2}\left( \cos \left( \dfrac{-\pi }{4} \right)+i\sin \left( \dfrac{-\pi }{4} \right) \right)$.
As, we know that $\cos \left( -\theta \right)=\cos \theta \text{ and }\sin \left( -\theta \right)=-\sin \theta $. By simplifying the above expression, the final answer will be \[Z=\sqrt{2}\left( \cos \dfrac{\pi }{4}-i\sin \dfrac{\pi }{4} \right)\]

Note: This problem could be alternatively solved by squaring and adding equation (3) and (4) to obtain the value of r. Students must be careful while equating both the equation and the argument must be determined considering the negative sign of the complex number.