What is $2z-\bar{z}$ if $\frac{\sqrt{2}}{z+i}=-|z|$?

Question

Let $z \in \Bbb{C}$. The following expression is true: $$\frac{\sqrt{2}}{z+i}=-|z|$$ Then, what is $$2z-\bar{z}$$ I'd really appreciate help on this one since I really don't have an idea what to do.

Answer 1

Notice $-|z|$ is a real number and $\sqrt{2}$ is a real number.

So $z + i= \frac {\sqrt 2}{-|z|}$ is a real number. So $z = a - i$ for some real $a$.

So $\frac {\sqrt{2}}{z + i}=-|z|$

$\frac {\sqrt 2}{a} = -\sqrt{a^2 + 1}$.

So $a < 0$ and $\sqrt 2 = -a\sqrt{a^2 + 1}$

$2 = a^2(a^2 + 1)$

$a^4 + a^2 -2 = 0$ so $a^2 = \frac {-1 \pm \sqrt{1+4*2}}2 = \frac {-1\pm 3}2$. As $a^2 \ge 0$, $a^2 = 1$ and as $a < 0$, $a = -1$ so

$z = -1 -i$

And $2z - \overline z = -2-2i - (-1 +i) = -1 - 3i$.

There maybe some other clever way to solve without solving $z$ first.

Answer 2

Note that the right-hand side of the given equation is a non-positive real therefore it is real also the left-hand side which implies that $z=x-i$ for some real number $x\leq 0$. Can you take it from here?

Answer 3

It follows that: $$\bigg|\frac{\sqrt2}{z+i}\bigg|=\bigg|-|z|\bigg|$$ And then $$\frac{\sqrt2}{|z+i|}=|z|$$ And so, taking $z=a+bi$ we get: $$\sqrt2=\sqrt{a^2+b^2}\cdot\sqrt{a^2+(b+1)^2}$$ See what you can do with this.

Also, $2z-\bar{z}=3a+bi$

Answer 4

Write: $$\underbrace{\frac{\sqrt{2}}{-|z|}}_{a\in \mathbb{R}}=z+i$$

so $z=a-i$. Now plug this in to a given equation:

$$ \sqrt{2} = -a\sqrt{a^2+1} \implies a^4+a^2-2=0 \implies a=\pm 1$$

Since $a<0$ we have $z=-1-i$. So $$2z-\overline{z} =-1-3i$$

Answer 5

HINT: You can use the fact that $$\left|z\right|^2=z\bar{z} \Rightarrow {\bar{z}=\frac{\left|z\right|^2}{z}}$$ And $$\left|z\right|^2=\frac{2}{\left(z+i\right)^2}$$ as stated in your original problem.

Answer 6

Hint:   the given equation can be rewritten as $\displaystyle z = - \frac{\sqrt{2}}{|z|}-i\,$, then:

$$\require{cancel} |z|^2 = z \cdot \overline z=\left(- \frac{\sqrt{2}}{|z|}-i\right)\cdot\left(-\frac{\sqrt{2}}{|z|}+i\right) = \frac{2}{|z|^2} - \cancel{i \frac{\sqrt{2}}{|z|}} + \cancel{i \frac{\sqrt{2}}{|z|}} +1$$

Therefore $\,|z|^4-|z|^2-2=0 \implies |z|^2 = 2\,$, and substituting $\,|z|=\sqrt{2}\,$ back into the original equation gives $\,z\,$.