Solve $z^2 - iz = |z - i|$

Question

I have the equation:

$z^2 - iz = |z - i|$

The solutions are $i$, $-\sqrt3/2 + i/2$, $\sqrt3/2 + i/2$

Can someone please walk me through or give me a hint...

Answer 1

A simple solution might be to try taking x and y such that z=x+iy and just plug through the algebra.

Answer 2

I'll try a more conceptual approach.

First thing you should notice is this: $$ z(z-i) = |z-i| $$ That should give you the trivial solution $z = i$. Next, recall that $\frac{z}{|z|} = e^{i \arg{z}}$. $$ \frac{z-i}{|z-i|} = \frac{1}{z} \\ e^{i \arg{(z-i)}} = \frac{1}{z} $$ It should now be quite clear that $|z| = 1$. Now we just need to figure out the argument/angle. Let $z = e^{i\arg{z}}$. $$ e^{i \arg{(z-i)}} = e^{-i\arg{z}} \\ \arg{(z-i)} = -\arg{z} $$ Graphically, what this means is that we want to find all of the angles such that the negative angle on the unit circle is exactly the same as subtracting $i$ (going down one unit) from the original angle. That should give you the other two solutions.

Answer 3

Let $z=x+iy$, $z^2-iz=|z-i|$ implies $0=\Im (z^2-iz)=2xy-x$ then $x=0$ or $y=\frac{1}{2}$.

If $x=0$, then we have $-y^2+y=|y-1|$ i.e. $-y(y-1)=|y-1|$ then $y=1$. So $z=i$ satisfies the equation.

Note that $z^2-iz=|z-i|$ implies $|z||z-i|=|z^2-iz|=|z-i|$, then $|z|=1$. So, if $y=\frac{1}{2}$, then $x=\pm\sqrt{1-\left(\frac{1}{2}\right)^2}=\pm\frac{\sqrt{3}}{2}$. If we put $z=\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}$ we get \begin{align} z^2-iz & =z(z-i) \\ & =\left(\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\left(\pm \frac{\sqrt{3}}{2}-i\frac{1}{2}\right) \\ & =\frac{3}{4}+\frac{1}{4} \\ & = 1 \\ \end{align} We can see easily $|z-i|=\left|\pm \frac{\sqrt{3}}{2}-i\frac{1}{2}\right|=1$.

Therefore, $i$, $\pm \frac{\sqrt{3}}{2}+i\frac{1}{2}$ are the roots.

Answer 4

Here is a solution which is little different, algebraically, though it is "spiritually similar" to the work of Mario Gallegos:

given that

$z^2 - iz = \vert z - i \vert, \tag{1}$

we first take the modulus of each side, noting that

$\vert z^2 - iz \vert = \vert z(z -i) \vert = \vert z \vert \vert z -i \vert; \tag{2}$

we thus obtain

$\vert z \vert \vert z -i\vert = \vert \vert z -i \vert \vert = \vert z -i\vert. \tag{3}$

From (3) we see that $\vert z -i \vert \ne 0$ implies $\vert z \vert = 1$; $z =i$ is clearly a solution as may be seen directly from (1); so we look for solutions with $\vert z \vert = 1$, $z \ne i$. $\vert z \vert = 1$ implies $z = e^{i\theta}$ with $0 \le \theta < 2\pi$; then

$z^2 - iz = e^{2i\theta} - ie^{i\theta} = \cos 2 \theta +i\sin 2\theta -i\cos \theta + \sin \theta. \tag{4}$

We see from (1) that $z^2 -iz$ must be real; this means the imaginary part of (4) must vanish, or

$\sin 2\theta - \cos \theta = 0; \tag{5}$

since $\sin 2 \theta = 2(\sin \theta)(\cos \theta)$ we have

$2(\sin \theta)(\cos \theta) = \cos \theta: \tag{6}$

if $\cos \theta = 0$, then $\theta = \pi / 2, 3\pi/2 \;$ or $z = \pm i$; we have seen that $z = i$ is a solution and $z = -i$ may be ruled out by direct substitution into (1); thus we may assume $\cos \theta \ne 0$, in which case we have $\sin \theta = 1/2$, or $\Im z = 1/2$. We now use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to arrive at $\Re z = \cos \theta = \pm \sqrt 3 / 2$, so that in addition to $z =i$, the other two solutions are

$z = \pm \dfrac{\sqrt 3}{2} + \dfrac{i}{2}; \tag{7}$

all three solution may be verified by direct substitution into (1).

Hope this helps. Cheerio,

and.as always,

Fiat Lux!!!