# Find $z\in\mathbb{C}$ such that $|z|=2i(\bar{z}+1)$

I was trying to solve the following problem:

Find $z\in\mathbb{C}$ such that $$|z|=2i(\bar{z}+1).$$

I tried to solve the problem by letting $z=x+iy$ and found out that the solutions are $z=-1+i\frac{1}{\sqrt{3}},-1-i\frac{1}{\sqrt{3}}$. But as tried to justify the solutions $z=-1-i\frac{1}{\sqrt{3}}$ does not satisfy the equation.

I couldn't find any error in my calculation. Any help would be highly appreciated. Thanks in advance.

– user261263
Aug 25 '17 at 7:00

Let $z=a+bi$, where $\{a,b\}\subset\mathbb R$.

Thus, $$\sqrt{a^2+b^2}=2i(a-bi+1),$$ which gives $a+1=0$ and $\sqrt{a^2+b^2}=2b$.

Hence, $\sqrt{b^2+1}=2b$ and since $b>0$, we get the answer: $$\left\{-1+\frac{1}{\sqrt3}i\right\}$$

Rewriting the equation:

$$|z|=2i(\bar{z}+1) \;\;\iff\;\; \bar{z} = \frac{|z|}{2i} - 1 = - 1 -i \frac{|z|}{2}\,$$

Taking conjugates on both sides:

$$z = - 1 + i \,\frac{|z|}{2} \tag{1}$$

Multiplying the two:

$$|z|^2= \left( - 1-i \frac{|z|}{2}\right) \left( - 1 + i \frac{|z|}{2}\right) = (-1)^2 - \left(i \frac{|z|}{2}\right)^2 = 1 + \frac{|z|^2}{4} \;\iff\; |z|^2 = 4/3$$

Substituting back in $(1)\,$:

$$z = - 1 + i \frac{|z|}{2} = -1 + i \,\frac{1}{2} \,\sqrt{\frac{4}{3}} = -1 + i \, \frac{1}{\sqrt{3}}$$

The equation you should get is

$$\sqrt{x^2+y^2} = 2i(x-iy + 1)\\ \sqrt{x^2 + y^2} = y + 2(x+1)i$$

which means that

1. $2y=\sqrt{x^2+y^2}$
2. $2(x+1)=0$

From the second equation, you get $x=-1$, and the first equation becomes

$$2y=\sqrt{y^2+1}$$

Now, you SQUARE the equation and get $$4y^2=y^2+1$$

and get $$y^2=\frac{1}{3}$$

Now, the equation $y^2=\frac13$ has two solutions, but since you got that equation by squaring some other equation, you should know that some of the solutions of the second equation may not solve the first equation. In your case, $y=\frac{1}{\sqrt{3}}$ solves the equation, because $$2\cdot \frac1{\sqrt{3}}=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+1}$$

but $y=-\frac{1}{\sqrt{3}}$ does not solve the equation, because $$2\cdot\left(- \frac1{\sqrt{3}}\right)\neq 2\cdot\frac1{\sqrt{3}}=\sqrt{\left(-\frac{1}{\sqrt{3}}\right)^2+1}$$

Here I think is your mistake: from $\sqrt{x^2+y^2}=2i(x-yi+1)$ you get $\sqrt{x^2+y^2}=2y$ therefore $y \ge 0$. So after finding out $x, y$ you must filter out the solutions that doesn't satisfy $y \ge 0$

Here is the solution by a different means... Let $z=re^{i\theta}$. Then

\begin{align} |z| &=r=2i(re^{-i\theta}+1)\\ &=2i(r\cos\theta-i\sin\theta+1) \end{align}

From the real and imaginary parts we have

$$0=r\cos\theta+1\\ r=2r\sin\theta$$

From the second equation we get $\theta=\pi/6$ or $\theta=5\pi/6$. However, we must choose $\theta=-5\pi/6$ to assure that $r$ is positive. Thus,

$$r=-\frac{1}{\cos\theta}=\frac{2}{\sqrt{3}}$$

and finally,

\begin{align} z &=re^{i\theta}\\ &=\frac{2}{\sqrt{3}e^{i5\pi/3}}\\ &=\frac{2}{\sqrt{3}}\left(-\frac{\sqrt{3}}{2}+\frac{i}{2}\right)\\ &=-1+\frac{i}{\sqrt{3}} \end{align}

in agreement with the other solutions.