Question 5 - Jee advanced Math 2022 P1 Questions with Solutions

Let \(\bar{z}\) denote the complex conjugate of a complex number \(z\) and let \(i=\sqrt{-1}\). In the set of complex numbers, the number of distinct roots of the equation 

\(\bar{z} - z^{2} = i(\bar{z} + z^{2})\)

is ______.

Sol:

\(\bar{z} - z^{2} = i(\bar{z} + z^{2})\)

\(\implies \bar{z}(1-i) = z^{2}(1+i)\)
Clearly \(z = 0 + 0i\) is one of the solutions of the equation. To find other non-zero solutions, let \(z \neq 0 + 0i \implies \bar{z} \neq 0+0i\) 

\(\implies \frac{z^{2}}{\bar{z}} = \frac{1-i}{1+i}\)

\(\implies \frac{z^{2}}{\bar{z}} = \frac{(1-i)(1-i)}{(1+i)(1-i)}\)

\(\implies \frac{z^{2}}{\bar{z}} = \frac{-2i}{2}\)

\(\implies \frac{z^{2}}{\bar{z}} = -i\)

Rewriting the above equation in exponential form,

\(\implies \frac{(r e^{i \theta})^{2}}{r e^{i (-\theta)}} = 1 e^{i (- \frac{\pi}{2})}\)
where \(r = |z|\) and \(\theta = arg(z)\)

\(\implies r e^{i (3\theta)} = 1 e^{i (- \frac{\pi}{2})}\)

Solutions to the equation are all the possible values of \(r\) and \(\theta\) for which the above equality holds.

We will use the theorem which states that if \(z_{1} = z_{2} \implies  r_{1} = r_{2}\) and \(arg(z_{1}) = arg(z_{2}) + 2\pi n\) where \(n = 0, \pm{1}, \pm{2},….\) 

\(\implies r = 1\) and \(3\theta = -\frac{\pi}{2} + 2\pi n\) where \(n  = 0, \pm{1}, \pm{2},….\)

\(\implies r = 1\) and \(\theta = \frac{\pi}{6}(4n - 1)\) where \(n  = 0, \pm{1}, \pm{2},….\)

Therefore, the solutions are of the form:

\(z = 1e^{i (\frac{\pi}{6}(4n - 1))}\) where \(n  = 0, \pm{1}, \pm{2},….\)

For \(n = 0 \implies z = 1e^{i (-\frac{\pi}{6})}\)
For \(n = 1 \implies z = 1e^{i \frac{\pi}{2}}\)
For \(n = 2 \implies z = 1e^{i \frac{7\pi}{6}}\)

For \(n = 3 \implies z = 1e^{i \frac{11\pi}{6}}\)
But this is same as \(z = 1e^{i(-\frac{\pi}{6}+2\pi)}\)    
\(\implies 1(\cos({-\frac{\pi}{6}+2\pi}) + i\sin({-\frac{\pi}{6}+2\pi}))\)
    \( =  \cos{(-\frac{\pi}{6}}) + i\sin({-\frac{\pi}{6}})\)
    \( = 1e^{i (-\frac{\pi}{6})}\)
So we end up at the same value of \(z\) where we began. 

(\implies\) for three distinct values of \(\theta\), there are 3 distinct non-zero roots. Also \(z = 0 + 0i\) makes it a total of 4 distinct roots to the equation. 
 

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