Question

Let z = a + ib, b≠ 0 be complex numbers satisfying

\(\begin{array}{l} z^2=\overline{z}\cdot 2^{1-\left|z\right|}.\end{array}\)

Then the least value of n∈ N, such that zⁿ= (z + 1)ⁿ, is equal to _____.

Answer 1

\(\begin{array}{l} \ z^2=\overline{z}\cdot 2^{1-\left|z\right|}\cdots\left(1\right) \end{array}\)

\(\begin{array}{l} \Rightarrow\ \left|z\right|^2=\left|\overline{z}\right|\cdot 2^{1-\left|z\right|}\end{array}\)

\(\begin{array}{l} \Rightarrow\ \left|z\right|=2^{1-\left|z\right|},\because\ b\neq0\Rightarrow \left|z\right|\neq 0 \end{array}\)

∴ |z| = 1 …(2)

\(\begin{array}{l}\because z = a + ib\ \text{then}\ \sqrt{a^2+b^2}=1 \cdots (3)\end{array}\)

Now again from equation (1), equation (2), equation (3) we get :

a² – b² + i2ab = (a – ib) 2⁰

∴ a² – b² =a and 2ab = – b

\(\begin{array}{l}\therefore \ a=-\frac{1}{2}\text{ and }b=\pm \frac{\sqrt{3}}{2}\end{array}\)

\(\begin{array}{l} \therefore\ z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\text{ or }z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array}\)

\(\begin{array}{l} z^n=\left(z+1\right)^n\Rightarrow \left(\frac{z+1}{z}\right)^n=1\end{array}\)

\(\begin{array}{l} \left(1+\frac{1}{z}\right)^n=1\end{array}\)

\(\begin{array}{l} \left(\frac{1+\sqrt{3}i}{2}\right)^n=1\end{array}\)

So, Minimum value of n is 6.