Given:
The complex number is
[tex]-1-i[/tex]
To find:
The polar form of given complex number.
Solution:
If z=a+ib, then
[tex]r=|z|=\sqrt{a^2+b^2}[/tex]
[tex]\tan \theta = \dfrac{b}{a}[/tex]
The polar form is [tex]z=r(\cos \theta +i\sin \theta)[/tex].
Let the complex number is
[tex]z=-1-i[/tex]
Here, a=-1, b=-1. Both a and b are negative. So, z lies in III quadrant.
[tex]r=\sqrt{(-1)^2+(-1)^2}[/tex]
[tex]r=\sqrt{1+1}[/tex]
[tex]r=\sqrt{2}[/tex]
And,
[tex]\tan \theta = \dfrac{-1}{-1}[/tex]
[tex]\tan \theta = 1[/tex]
[tex]\tan \theta = \tan (\pi+\dfrac{\pi}{4})[/tex] (because z lies in III quadrant)
[tex]\tan \theta = \tan (\dfrac{5\pi}{4})[/tex]
[tex]\theta = \dfrac{5\pi}{4}[/tex]
Now, the polar form of given complex number is
[tex]z=r(\cos \theta +i\sin \theta)[/tex]
[tex]z=\sqrt{2}(\cos \dfrac{5\pi}{4} +i\sin \dfrac{5\pi}{4})[/tex]
Therefore, the required polar form is [tex]z=\sqrt{2}(\cos \dfrac{5\pi}{4} +i\sin \dfrac{5\pi}{4})[/tex].
