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Question:

If $z \neq 1$ and ${z^2} \over {z-1}$ is real, then find the locus of the point represented by the complex number $z$.

I'm not sure how to approach this question. I attempted to substitute $z = x + iy$, however, that didn't solve the problem. It's quite clear, by observation, that any point on the real axis would satisfy the equation. However, there also have to be some other points. How would I approach this question?

Gummy bears
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3 Answers3

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$$\dfrac{z^2}{z-1}=z+1+\dfrac1{z-1}$$ will be real iff $z-1+\dfrac1{z-1}$ is real

Set $z-1=w$

Now $w+\dfrac1w$ is real $\iff w+\dfrac1w=\bar w+\dfrac1{\bar w}$

$(w-\bar w)\dfrac{w\bar w-1}{w\bar w}=0$

If $w-\bar w=0, w=\bar w\implies w$ is real $\iff z$ is real

Else $w\bar w-1=0\iff1=w\bar w=|w|^2$

lab bhattacharjee
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Let $\displaystyle z=x+iy$ in $\displaystyle \frac{z^2}{z-1} = \frac{(x+iy)^2}{x+iy-1} = \frac{x^2-y^2+2ixy}{(x-1)+iy}\times \frac{(x-1)-iy}{(x-1)-iy}$

So we get $\displaystyle \frac{(x^2-y^2)(x-1)+2xy^2+i\left[-y(x^2-y^2)++2xy(x-1)\right]}{(x-1)^2+y^2}$

$\displaystyle = \frac{(x^2-y^2)(x-1)+2xy^2}{(x-1)^2+y^2}+i\frac{2xy(x-1)-y(x^2-y^2)}{(x-1)^2+y^2}$

so put $\displaystyle \frac{2xy(x-1)-y(x^2-y^2)}{(x-1)^2+y^2} = 0\Rightarrow 2xy(x-1)-y(x^2-y^2) =0$

So we get $\displaystyle y\left[2x^2-2x-x^2+y^2\right] = 0\Rightarrow y[x^2+y^2-2x] =0$

So $y=0$ or $x^2+y^2-2x=0$

juantheron
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Using If ${z'}$ is real Then $z'=\bar{z'}$

So Let $$\displaystyle z'=\frac{z^2}{z-1}\;,$$ Then $$\displaystyle \frac{z^2}{z-1} = \upperbrace{\left(\frac{z^2}{z-1}\right)} = \frac{\bar{z}^2}{\bar{z}-1}$$

So we get $$\displaystyle \displaystyle \frac{z^2}{z-1} = \frac{\bar{z}^2}{\bar{z}-1}$$

So $$\displaystyle z^2\bar{z}-z^2 = \bar{z}^2z-\bar{z}^2\Rightarrow \left(z^2\bar{z}-z\bar{z}^2\right)-(z^2-\bar{z}^2) =0$$

so we get $$\displaystyle z\bar{z}(z-\bar{z})-(z+\bar{z})\cdot (z-\bar{z}) =0$$

So we get $$\displaystyle (z-\bar{z})\cdot \left[z\bar{z}-(z+\bar{z})\right] =0$$

So we get $$z-\bar{z} =0$$ or $z\bar{z}-(z+\bar{z})=0$$

Now Put $z=x+iy$ and $\bar{z} = x-iy\;,$ means $z+\bar{z} = 2x$ and $z-\bar{z}=2iy$

and $z\bar{z} = x^2+y^2$

So we get $$2iy=0\Rightarrow y=0$$ or $$x^2+y^2-2x =0$$

juantheron
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