Prove that $z - \frac{4}{z}$ is purely imaginary if and only if z is purely imaginary or |z|= 2.

Question

I am trying to prove that for any $z \neq 0$, $z - \frac{4}{z}$ is purely imaginary if and only if z is purely imaginary or |z|= 2.

I tried writing = + and since z is purely imaginary, $x=0$ and $z - \frac{4}{z}$ = $iy - \frac{4}{iy}$. I don’t know if this makes sense.

Any help is appreciated! Thank you!!

Answer 1

Let $z=x +i y$,

$$w=z-\frac4z = x+iy-\frac4{x+iy} = x+iy-\frac{4(x-iy)}{|z|^2}$$ $$= \left( 1-\frac 4{|z|^2}\right)x +i \left( 1+\frac 4{|z|^2}\right)y$$

Now, you could argue both ways.

$$x=0\>\>or \>\> |z|=2 \iff Re(w) =0$$

Answer 2

We can use that $2\Re(w)=w+\bar w$ therefore

$$2\Re\left(z - \frac{4}{z}\right)=z - \frac{4}{z}+\bar z - \frac{4}{\bar z}=2\Re(z)-8\frac{\Re (z)}{z\bar z}=2\Re(z)\left(1-\frac4{|z|^2}\right)=0 $$

$$\iff \Re(z)=0 \quad \lor\quad |z|^2=4$$

Answer 3

$z\overline{z}=|z|^2\Rightarrow z-4/z=z-4\overline{z}/|z|^2=\begin{cases} z-\overline{z}=2iIm(z) \text{ if } |z|=2 \\ z+4z/|z|^2=z(1+4/|z|^2)\in i\mathbb{R}\text{ if }z\in i\mathbb{R} \end{cases}$

Answer 4

$z = x+iy\\ z - \frac {4}{z} = x+iy - \frac {4x - 4iy}{x^2 + y^2}\\ z - \frac {4}{z} = x - \frac {4x}{x^2+y^2} +i(y + \frac {4y}{x^2+y^2})$

$z - \frac {4}{z}$ is pure imaginary $\iff x - \frac {4x}{|z|^2} = 0$

Answer 5

So exists $r \in\mathbb{R}$ such that $$z-{4\over z} =ir$$ thus $$z^2-irz-4=0$$

Disciriminat is $D= -r^2+16$ so $$z = {ir+\sqrt{D}\over 2}$$

So if $D<0$ then $z$ is imaginary and if $D\geq 0$ then $$|z| = {1\over 2}\sqrt{r^2+D}= 2$$

Answer 6

Let $z=x+iy$

$\dfrac rz=\dfrac r{x+iy}=\dfrac{r(x-iy)}{x^2+y^2}$

We need $x-\dfrac{rx}{x^2+y^2}=0$

$\implies x\left(x^2+y^2-r\right)=0$

Here $r=4$