Let z and $\omega $ be two complex numbers such that $\omega  = z\bar z – 2z + 2$, $\left| {\frac{{z + i}}{{z – 3i}}} \right| = 1$ and $Re(\omega )$ has minimum value. Then, the minimum value of $n \in N $ for which $\omega ^n $ is real, is equal to _ _ _ _ .

Solution

Let z = x + iy

We have, $\left| {\frac{{x + iy + i}}{{x + iy – 3i}}} \right| = 1$

$ \Rightarrow |x + i(y + 1)| = |x + i(y – 3)|$

$ \Rightarrow {x^2} + {(y + 1)^2} = {x^2} + {(y – 3)^2}$

$ \Rightarrow 2y + 1 =  – 6y + 9$

$ \Rightarrow y = 1$

So, $z = x + i$

Now, $\omega  = z\bar z – 2z + 2 = {x^2} + {1^2} – 2(x + i) + 2 = {x^2} – 2x + 3 – 2i$

${\mathop{\rm Re}\nolimits} (\omega ) = {x^2} – 2x + 3 = {(x – 1)^2} + 2 = 2 + non – negative$

${\{ {\mathop{\rm Re}\nolimits} (\omega )\} _{min.}} = 2$

$\omega  = 2 – 2i = 2\sqrt 2 \left( {\frac{1}{{\sqrt 2 }} – \frac{1}{{\sqrt 2 }}i} \right) = 2\sqrt 2 \left[ {\cos ( – \frac{\pi }{4}) + i\sin (- \frac{\pi }{4}}) \right]$

${\omega ^n} = {(2\sqrt 2 )^n}\left[ {\cos (- \frac{{n\pi }}{4} ) + i\sin (- \frac{{n\pi }}{4}}) \right]$

For $\omega ^n $ to be real, $\sin ( – \frac{{n\pi }}{4} ) = 0$

The smallest natural number n satisfying above = 4