ii) De Moivre’s Theorum
Let \(\displaystyle \L z \, = \, 1 \, + \, i\)
Although this is relatively ok, it's always a good idea to sketch an Argand diagram to ensure we calculate the correct argument.
Code:
/|\ Im
|
|
|
1 + - - - - *
| * :
| * :
| *\ :
| * | :
--------+---------+----->
| 1 Re
|
|
\(\displaystyle \L |z| \, = \sqrt{1^2 \, + \, 1^2} \, = \, \sqrt{2} \, = \, 2^{\frac{1}{2}}\)
\(\displaystyle \L Arg(z) \, = \, \tan^{-1}{\left(\frac{ \, 1 \, }{ \, 1 \, }\right)} \, = \, \frac{\pi}{4}\)
So \(\displaystyle \L z \, = \, 2^{\frac{1}{2}} \left(\cos{\left(\frac{\pi}{4}\right)} \, + \, i\sin{\left(\frac{\pi}{4}\right)}\right)\)
Now apply De Moivre's Theorem to find \(\displaystyle \L z^{\frac{ \, 1 \, }{ \, 2 \, }}\) .