Notes: Write the complex number `z=-1-i` in polar form.

Friday, 19 August 2016

Write the complex number $\displaystyle{z}=-{1}-{i}$ in polar form.

We are asked to write the complex number z=-1-i in polar form.

The polar form of a complex number is

$\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}$

where r is the distance from the origin (the modulus or absolute value of the complex number) and theta is the angle from the positive x-axis. (Note that the angle is not unique -- you can always select an angle from 0 to 2pi or 0 to 360 degrees.)

To compute r we use

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

where a is the real part and b is the imaginary part of the complex number. So:

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

We can find theta by

$\displaystyle{\tan{\theta}}=\frac{{b}}{{a}}$

In this case we can solve the angle by inspection as 225 degrees or 5pi/4.

The polar form is:

$\displaystyle{z}=\sqrt{{{2}}}{\left({\cos{{\left(\frac{{{5}\pi}}{{4}}\right)}}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{4}}\right)}}}\right)}$

An alternative is to write in Euler notation:

$\displaystyle{z}={\left|{z}\right|}{{e}}^{{{i}\theta}}={r}{{e}}^{{{i}\theta}}$

So here we have:

$\displaystyle-{1}-{i}=\sqrt{{{2}}}{{e}}^{{{\left({i}\frac{{{5}\pi}}{{4}}\right)}}}$

Note again that the angle is not unique; we can add/subtract any multiples of 2pi and still have the same point. A general solution is:

$\displaystyle-{1}-{i}=\sqrt{{{2}}}{\left({\cos{{\left(\frac{{{5}\pi}}{{4}}+{2}\pi{n}\right)}}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{4}}+{2}\pi{n}\right)}}}\right)}=\sqrt{{{2}}}{{e}}^{{{i}{\left(\frac{{{5}\pi}}{{4}}+{2}\pi{n}\right)}}};{n}\in\mathbb{Z}$