A non-zero complex number $z$ satisfies $\vert \frac{z}{z+1}\vert=1$ and $\frac{z}{\bar z}=i$.Find $z?$

Question

A non-zero complex number $z$ satisfies $\vert \frac{z}{z+1}\vert=1$ and $\frac{z}{\bar z}=i$.Find $z?$

I know that $\vert \frac{z}{z+1}\vert=1$ represent an ellipse with focii at $(0,0)$ and $(-1,0)$.But I do not know how solve $\vert \frac{z}{z+1}\vert=1$ and $\frac{z}{\bar z}=i$ simultaneously.

Need Help!!

Answer 1

Start with first expression, and with $z=x+yi$: $$\lvert \frac{z}{z+1} \rvert= 1 \iff \lvert z \rvert = \lvert z+1 \rvert \iff \sqrt{x^2+y^2} = \sqrt{(x+1)^2+y^2} \\ \iff x^2+y^2=x^2+2x+1+y^2\\0 = 2x+1 \iff x=-\frac{1}{2}.$$

Now, proceed with second expression: $$\frac{z}{\overline z} =i \iff z=\overline zi \iff x+yi=xi+y.$$

The last equallity is true if and only if $\quad x=y, \quad$ so it will be: $$z=-\frac{1}{2}-\frac{1}{2}i.$$

Answer 2

If $z=x+iy$ then $\frac{z}{\bar{z}} = i$ means $\dfrac{x^2-y^2+2xyi}{x^2+y^2} = i$, i.e. $x^2=y^2$ and $2xy=x^2+y^2$, which is equal to $2x^2$ from the first equation.

You can also show $|z|=|z+1|$ gives $x^2 = (x+1)^2$.

Answer 3

Let $z=x+iy$.

• $\left|\frac{z}{z+1}\right|=1$ means $|z|=|z+1|$, i.e. $z$ is equidistant from the points $0$ and $-1$. This gives you $x=-\frac{1}{2}$ (with arbitrary $y$).

• $\frac{z}{\bar{z}}=i$ means $z=i\bar{z}$, i.e. $x+iy=i(x-iy)=y+ix$, which gives you $x=y$.

Both of those together give you $x=y=-\frac{1}{2}$, i.e. $z=-\frac{1}{2}-i\frac{1}{2}$.