Find the roots of $\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}$

Question

I attempted the problem in two ways

Method 1
Resolved $\mathrm{ Z = X + iY}$ this led to a very big cubic equation and handling that became too difficult for me .

Method 2
I believe factorizing the equation is a far better method, I don't know how to go about it .

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EDIT
The answer given is $\mathrm{Z = \frac{-i}{2} , i}$

As pointed out by @gimsui $\mathrm{\frac{-i}{2} }$ is not a solution most likely a printing error.

Answer 1

The given equation is equivalent to $Z^3=-i$, indeed note simply that

$$\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}\iff\bar Z-iZ^2=|Z|^2-1$$

then $\bar Z-iZ^2$ is real which requires

• -$iZ^2=Z\implies Z=0 \quad \lor \quad -iZ=1\implies Z=i$ ($Z=0$ is not a solution)

or

• $\bar Z-iZ^2=0\implies$ (from original equation) $|Z|=1$

thus the condition $\bar Z-iZ^2$ is real requires $|Z|=1$ and the solution is given by

$$\bar Z-iZ^2=0\implies Z^3=-i \implies Z_1=i,\,Z_{2,3}=\frac{-i\pm\sqrt 3}{2}$$

Answer 2

The complex number $Z$ can be written as $$Z = A + iB.$$

Then, your equation becomes:

$$\overline{Z} + 1 = iZ^2 + |Z|^2 \Rightarrow \\ A-iB + 1 = i(A^2-B^2+2iAB)+A^2+B^2 \Rightarrow \\ A-iB + 1 = iA^2 - iB^2 -2AB + A^2 + B^2 \Rightarrow \\ (A+1 +2AB-A^2-B^2) + i(-B-A^2+B^2) = 0.$$

This means that both the real part $(A+1 +2AB-A^2-B^2)$ and the imaginary part $(-B-A^2+B^2)$ must be $0$. Then, you must solve the following system:

$$\begin{cases} A+1 +2AB-A^2-B^2 & = 0\\ -B-A^2+B^2 & = 0 \end{cases}. $$

This system has the following solutions:

$$A_1 = -\frac{\sqrt{3}}{2}, B_1 = -\frac{1}{2} \Rightarrow Z_1 = -\frac{\sqrt{3}}{2} -i\frac{1}{2},\\ A_2 = \frac{\sqrt{3}}{2}, B_2 = -\frac{1}{2} \Rightarrow Z_2 = \frac{\sqrt{3}}{2}-i\frac{1}{2}, \\ A_3 = 0, B_3 = 1 \Rightarrow Z_3 = i.$$

Answer 3

\begin{align} z &= x + iy \\ \hline \overline z &= x - iy \\ z^2 &= (x^2-y^2)+2ixy \\ |z|^2 &= x^2+y^2 \end{align}

\begin{align} \overline z + 1 &= iz^2 + |z|^2 \\ (x-iy) + 1 &= i(x^2-y^2 + 2ixy)+(x^2+y^2) \\ (x+1) - iy &= (x^2-2xy+y^2) + i(x^2-y^2) \\ x+1-x^2+2xy-y^2 &= i(x^2-y^2+y) \\ \hline x^2-2xy+y^2 - x - 1 &= 0 \\ x^2 &= y^2-y \\ \hline 2y^2-2xy-x-y-1 &= 0 \\ 2y^2 -(2x+1)y-(x+1)&=0 \\ y &= \dfrac{(2x+1)\pm \sqrt{(2x+1)^2+8(x+1)}}{4} \\ y &= \dfrac{(2x+1)\pm \sqrt{4x^2+12x+9}}{4} \\ y &= \dfrac{(2x+1)\pm |2x+3|}{4} \\ y &= \dfrac{(2x+1)\pm (2x+3)}{4} \\ y &\in \left\{ x+1, -\frac 12 \right\} \end{align}

If $y=x+1$ and $x^2=y^2-y$, then \begin{align} x^2 &= (x+1)^2 - (x+1) \\ x^2 &= x^2 + x \\ (x,y) &= (0,1) \end{align}

If $y=-\dfrac 12$ and $x^2=y^2-y$, then \begin{align} x^2 &= \dfrac 14 + \dfrac 12 \\ (x,y) &= \left( \pm \dfrac{\sqrt 3}{2}, -\dfrac 12 \right) \end{align}

$$z \in \left\{ i, -\dfrac{\sqrt 3}{2} - i\dfrac 12, \dfrac{\sqrt 3}{2} - i\dfrac 12 \right\} $$