Need help with solving these two equations:
(2z + i)[FONT="]^[/FONT]4 − 16i = 0
|z − 2 + i| + i ¯z = 3
The first one: I chose t as a substitution for (2z + i), that makes : t[FONT="]^[/FONT]4 = 16i , which is 90°. I dont know how to find the four solutions to this equation
The second one: substitution z = x + iy, for ¯z = x - yi, but how to continue?
Thanks.
To find the nth root of a complex number, in the form \(\displaystyle re^{i\theta}\), take the nth root of r and divide \(\displaystyle \theta\) by n (That is, basically D'Moivre's theorem that Subhotosh Khan referred to). 16i has r= 16 and \(\displaystyle \theta= \pi/2\). The fourth root of 16 is 2. \(\displaystyle \theta/4= \pi/8\) so that one solution is \(\displaystyle 2e^{i\pi/8}\)
However, if we take \(\displaystyle \theta= \frac{\pi}{2}+2\pi= \frac{5\pi}{2}\) we get the same value for \(\displaystyle e^{i\theta}\) but a different value for \(\displaystyle e^{i\theta/8}\). Another root is \(\displaystyle 2e^{5i\pi/8}\).
And we can do that again: \(\displaystyle \theta+ 4\pi=\)\(\displaystyle \frac{\pi}{2}+ \frac{8\pi}{2}= \)\(\displaystyle \frac{9\pi}{2}\) so a third root is \(\displaystyle 2e^{9i\pi/8}\).
One more time: \(\displaystyle \theta+ 6\pi= \frac{\pi/}{2}+ \frac{12\pi}{2}=\)\(\displaystyle \frac{13\pi}{2}\) so a fourth root is \(\displaystyle 2e^{12i\pi/8}\).
I will leave it to you to show that if we use \(\displaystyle \theta+ 8\pi= \frac{\pi}{2}+ \frac{16\pi}{2}= \frac{17\pi}{2}\) we get the first solution again.
For the second problem, setting z= x+ iy, we get \(\displaystyle |x+ iy- 2+ i|+ ix- iy= 3\). The first thing you should recognize is that since the right side of the equation, 3, is a real number, the left side must be also. The absolute value is a real number so we must have ix- iy= i(x- y)= 0. That is, y= x so we can write the equation as \(\displaystyle |(x- 2)+ i(x+ 1)|= \sqrt{(x- 2)^2+ (x+ 1)^2}= 3\) so that (x- 2)^2+ (x+ 1)^2= 2x^2- 2x+ 5= 9[/tex]. Solve the quadratic equation \(\displaystyle 2x^2- 2x- 4= 0\).
