Complex numbers - equations: (2z + i)^4 − 16i = 0, |z − 2 + i| + i ¯z = 3

[Isiltir]

Complex numbers - equations: (2z + i)^4 − 16i = 0, |z − 2 + i| + i ¯z = 3

Need help with solving these two equations:

(2z + i)[FONT=&quot]^[/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=&quot]^[/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.

 

[Deleted member 4993]

Need help with solving these two equations:

(2z + i)^4 − 16i = 0

|z − 2 + i| + i ¯z = 3

The first one: I chose t as a substitution for (2z + i), that makes : t^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.

Hint:

use DeMoivre's Theorem

 

[HallsofIvy]

Need help with solving these two equations:

(2z + i)[FONT=&quot]^[/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=&quot]^[/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 $\displaystyle 2x^2- 2x- 4= 0$.

 

[Isiltir]

But is that right?

|z − 2 + i| + i ¯z = 3

|z − 2 + i| + i. (x - iy) = 3

|z − 2 + i| + ix + y = 3


This would make i¯z =(ix +y) not (ix+iy)