Find all complex numbers z, which satisfy z^(2)= i - 1.
I have come across this question during my tutorial and I cant seem to solve it. I need some reference. Thank you---> to the kind soul who is able to help me out here... =D
God Bless.
Find all complex numbers z, which satisfy z^(2)= i - 1.
I have come across this question during my tutorial and I cant seem to solve it. I need some reference. Thank you---> to the kind soul who is able to help me out here... =D
God Bless.
$\displaystyle \displaystyle |z|=\sqrt{2} \ arg(z)=tan^{-1}\frac{y}{x}=tan^{-1}(-1)=\frac{3\pi}{4}$
$\displaystyle \displaystyle k=0,1 \ n=2 \ w_k=r^{1/n}*\left[cos\left(\frac{\theta+2\pi k}{n}\right)+isin\left(\frac{\theta+2\pi k}{n}\right)\right]$
$\displaystyle \displaysyle w_0=\sqrt{2}^{1/2}*\left[cos\left(\frac{\theta+2\pi *0}{2}\right)+isin\left(\frac{\theta+2\pi *0}{2}\right)\right]\cdots$
$\displaystyle \displaystyle w_1=\cdots$
Exponential equations are easiest to solve when you put the complex numbers in exponential form. Remember that the are always 2 square roots, 3 cube roots, 4 fourth roots, etc, and they are all evenly spaced around a circle.
So for square roots, they are separated by an angle of $\displaystyle \displaystyle \frac{2\pi}{2} = \pi$.
Anyway, $\displaystyle \displaystyle |-1 + i| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.
$\displaystyle \displaystyle \arg{(-1 + i)} = \pi - \arctan{\frac{1}{1}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
So that means $\displaystyle \displaystyle -1 + i = \sqrt{2}\,e^{\frac{3\pi i}{4}}$.
Therefore, $\displaystyle \displaystyle z^2 = \sqrt{2}\,e^{\frac{3\pi}{4}}$
$\displaystyle \displaystyle z = \left(\sqrt{2}\,e^{\frac{3\pi}{4}}\right)^{\frac{1 }{2}}$
$\displaystyle \displaystyle z = \sqrt[4]{2}\, e^{\frac{3\pi}{8}}$.
That is the first square root. The second has an extra $\displaystyle \displaystyle \pi$ added to its angle. So the other square root is
$\displaystyle \displaystyle z = \sqrt[4]{2}\,e^{\frac{11\pi}{8}}$.