How do I find the complex solution of this absolute value problem?

Question

I have the equation $|z-i|^4=1$, and the solution needs to be in the form of $x+iy$ or $re^{i\theta}$. First I attempted this: $$|x+iy-i|^4=1$$ $$|x+i(y-1)|=1$$ $$|z|=\sqrt{x^2+y^2} \ \Rightarrow \sqrt{x^2-(y-1)^2}=1$$ $$x^2+(y-1)^2=1$$ This is a circle centered at $1$ with a radius of $1$ which makes sense. However it's not in the correct form of $x+iy$ or $re^{i\theta}$, so I thought that maybe I could do some substitution to make it so. $$z=x+iy, \ \ \ w=-i$$ $$|z+w|=1$$ $$|z|+|w|=1 $$ $$|x+iy|+|-i|=1$$ $$|x+iy|+\sqrt{-i^2}=1$$ $$|x+iy|+1=1$$ $$z=x+iy=0$$

This puts it in the right format, but I'm not sure that it makes any sense. This says that $z=0$ which isn't a circle, and the other solution shows that it should indeed be a circle. Is it possible to put the solution to the first attempt in one of the forms that it's supposed to be in?

Answer 1

If $|z-i|^4=1 \Rightarrow |z-i|=1$, which implies $z-i=e^{i\theta}$ for some arbitrary angle $\theta$. This implies:

$$ z=i+e^{i\theta} = \cos\theta +i(1+\sin\theta)$$

Answer 2

Think about what $|z-i|^4 = 1$ means. It means that the magnitude of the complex number $z-i$, when raised to the fourth power, equals $1$. Since magnitude is always a nonnegative real number, it follows that the magnitude itself must equal $1$, since the only nonnegative real solution to the equation $$x^4 = 1$$ is $x = 1$. Therefore, the set of $z$ that satisfy $|z - i|^4 = 1$ is the same as the set that satisfy $|z-i| = 1$. This is obviously a circle of radius $1$ centered at $i$.

Now that we understand the desired locus, all that remains is to express it algebraically. To this end, we recall that the Cartesian equation of a circle centered at $(0,1)$ with radius $1$ is simply $$(x-0)^2 + (y-1)^2 = 1$$ or $$x^2 + y^2 - 2y = 0.$$ So in rectangular form, $z$ is characterized by $$x + iy = \pm \sqrt{2y - y^2} + iy, \quad y \in [0,1],$$ and here $y$ represents the imaginary part of $z$, geometrically the vertical distance from the real axis.

In polar form, let $z = re^{i\theta}$. We seek a relationship between $r$ and $\theta$ that describes such a circle. But since $e^{i\theta} = \cos\theta + i \sin \theta$, we simply have $$(r \cos \theta)^2 + (r \sin \theta)^2 - 2(r \sin \theta) = 0,$$ or $$r^2 - 2r \sin \theta = 0,$$ or $$r = 2 \sin \theta.$$ It follows that the desired locus is $$z = 2e^{i\theta} \sin\theta, \quad \theta \in [0,\pi),$$ noting that the circle is swept out twice if we were to use $\theta \in [0,2\pi)$. Here, $\theta$ represents the counterclockwise angle $z$ makes with the positive real axis.