If $\frac{w-\bar w z}{1-z}$ is purely real, where $w=\alpha +i\beta$, $\beta \ne 0$ and $z\ne 1$, then find set of values of $z$

Question

$$\frac{\alpha +i\beta-(\alpha-i\beta)z}{1-z}$$ Since it’s purely real $$\beta \frac {1+z}{1-z}=0$$

$$\beta \ne 0$$

$$\frac{1+z}{1-z}=0$$

Clearly $z\ne 1$. How do I find the other conditions?

Answer 1

We don't know whether $z$ is purely real or not. So you can't ignore $\alpha z$

$$\Rightarrow\frac{\alpha+i\beta-(\alpha-i\beta)z}{1-z} = \frac{\alpha(1-z) +i\beta(1+z)}{1-z} = \alpha+i\beta\frac{1+z}{1-z}$$

Let $z = x+iy$

$$\Rightarrow \alpha + i\beta\frac{(1+x)+iy}{(1-x)-iy} = \alpha+i\beta\frac{((1+x)+iy)((1-x)+iy)}{(1-x)^2+y^2} = \alpha+i\beta\frac{1-x^2-y^2+2yi}{(1-x)^2+y^2} = \alpha + \frac{\color{blue}{i\beta(1-x^2-y^2)}-2y\beta}{(1-x)^2+y^2}$$

Now equate the imaginary part to zero.

$$\Rightarrow x^2+y^2 = 1 $$

But $z\ne 1 \Rightarrow x \ne1,y\ne0$

So, $z = x+iy$ represents a unit circle which has a hole at $(1,0)$

Answer 2

Substitute $w=\frac{w+\bar{w}}{2}+\frac{w-\bar{w}} {2}\rightarrow\bar{w}=\frac{w+\bar{w}}{2}-\frac{w-\bar{w}}{2}$ to the equation

$$ \frac{w-\bar{w}z}{1-z}=\frac{w+\bar{w}}{2}+\frac{w-\bar{w}}{2}\frac{1+z}{1-z} $$

$\frac{1+z}{1-z}$ is pure imaginary if $(1,0)$, $z$, $(-1,0)$ form a right triangle with right angle at $z$. Therefore $z$ lies on a circle with $(1,0)$ and $(-1,0)$ as its diameter.

to conclude, $z$ lies on unit circle centered on origin except $(1,0)$

Answer 3

Simplify the expression $$\dfrac{w-\bar w z}{1-z}=\overline{\left(\dfrac{w-\bar w z}{1-z}\right)}.$$ It tells you either $w=\bar w$ or $|z|=1.$