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If $\frac{z-a}{z+a},a\in \mathbb{R}$ is a purely imaginary number and $|z|=2$, then a value of a is ___. (JEE Main 2019)

For a purely imaginary number $w$, we must have $w+\bar{w}=0$, giving us $a=2$. (This solution is standard substitution and simplification, resulting in a quadratic).

I tried it using a different method(below): $$\frac{z-a}{z+a}\times\frac{\bar{z}-a}{\bar{z}-a}=\frac{z\bar{z}-az-a\bar{z}+a^2}{z\bar{z}-a^2}=\frac{z\bar{z}-az-a\bar{z}+a^2}{|z|^2-a^2}=\frac{z\bar{z}-az-a\bar{z}+a^2}{4-a^2}$$ where in the last step we substituted $z\bar{z}=|z|^2=4, z+\bar{z}=0$$

The denominator of this expression (from the other, valid solution is zero), and hence the expression is not defined.

What was the mistake and at what step was it made?

Starlight
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2 Answers2

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You expanded $(z + a)(\bar{z} - a)$ as $z\bar{z} - a^2$, when the correct expression is $z\bar{z} + a\bar{z} - az - a^2$. As the comments point out, you also wrote that $z + \bar{z} = 0$, which you can't assume, since $z$ is not necessarily purely imaginary.

In general, for problems like this, pick a value of $z$ (any value) that fits your hypothesis, and plug it in at each step; you'll see where the error happens quite easily.

Nicholas Priebe
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$z+\bar z \neq 0$, unless, $\bar z = -z$ (i.e. $e^{i(-Arg(z))}=e^{i(Arg(z)-\pi)}$) or in other words unless ${\frak{Re}}(z)=0$), and

$(z+a)(\bar z -a)=z \bar z+a\bar z-az-a^2\neq z \bar z-a^2$, unless $z = \bar z$ or in other words unless ${\frak{Im}}(z)=0$

Indeed the canonical multiplier for removing the imaginary part of the denominator is $\frac{\bar z + a}{\bar z + a}$

So, $$\frac{z -a}{z +a}\frac{\bar z +a}{\bar z +a}=\frac{z \bar z+az-a\bar z-a^2}{z \bar z+az+a\bar z+a^2}=\frac{z \bar z+az-a\bar z-a^2}{|z+a|^2}$$

since the purpose of your choice of multiplier was to remove the imaginary part of the denominator by undoing the rotation ($e^{iArg(z)}$) by multiplying by a number with the reverse of its argument (i.e. by $e^{-iArg(z)}$) and by taking the fast route of re-using the number you already have, in conjugate, you are multiplying the magnitude by the magnitude and thus removing the rotation but squaring the magnitude.

Localth
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