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I am struggling with this question, it states:

z is complex such that\begin{align} & \frac{z}{z-i} \\\\ \end{align} is real.

Show that z is imaginary.


What is the best way to go about the question?

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  • $\begingroup$ @msm If I were to guess, that (plus the 3 votes to close) could well be because of the complete lack of context and effort on OP's part. For the record, I cast none of the above, but I find them entirely justified. $\endgroup$ – dxiv Oct 28 '16 at 4:18
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Hint: First off, rewrite it so that the denominator is real, i.e: $$\frac{z}{z-i}=\frac{z\cdot \overline{z-i}}{(z-i)\overline{z-i}}=\frac{z\cdot \overline{z-i}}{|z-i|^2},$$ so that your task is equivalent to showing that $$z\cdot \overline{z-i}=z\bar{z}+iz=|z|^2+iz$$ is real. But this is easy...

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With $r\in \mathbb{R}$ we have $$\frac{z}{z-i} = r$$ $$z=r(z-i)= rz - ri$$ $$z(1-r)=-ri$$ $$z = \frac{r}{r-1}i$$ So $z$ is a real multiple of $i,\;$ i.e. $z$ is imaginary.

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$w \in\mathbb{C}$ is real if and only if $w = \overline w$. Applying that to $w=\frac{z}{z-i}$ gives:

$$ \begin{align} \frac{z}{z-i} = \frac{\overline z}{\overline z+i} & \iff z(\overline z + i) = \overline z(z-i) \iff z \overline z + i z = z \overline z - i \overline z \iff z = - \overline z \\ & \iff \operatorname{Re}(z) = \frac{1}{2}(z+\overline z) = 0 \;\iff\; z \in i\, \mathbb{R} \end{align} $$

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We need to solve, when $\text{z}\in\mathbb{C}$:

$$\Im\left[\frac{\text{z}}{\text{z}-i}\right]=0$$

So, we get:

  1. $$\Re\left[\frac{\text{z}}{\text{z}-i}\right]=\Re\left[\frac{\Re\left[\text{z}\right]+\Im\left[\text{z}\right]i}{\Re\left[\text{z}\right]+\Im\left[\text{z}\right]i-i}\right]=1+\frac{\Im\left[\text{z}\right]-1}{\Re^2\left[\text{z}\right]+\left(\Im\left[\text{z}\right]-1\right)^2}$$
  2. $$\Im\left[\frac{\text{z}}{\text{z}-i}\right]=\Im\left[\frac{\Re\left[\text{z}\right]+\Im\left[\text{z}\right]i}{\Re\left[\text{z}\right]+\Im\left[\text{z}\right]i-i}\right]=\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\left(\Im\left[\text{z}\right]-1\right)^2}$$

So:

$$\Im\left[\frac{\text{z}}{\text{z}-i}\right]=0\Longleftrightarrow\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\left(\Im\left[\text{z}\right]-1\right)^2}=0\Longleftrightarrow\Re\left[\text{z}\right]=0$$

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