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 z n= (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 :

a 2 – b 2 + i 2 ab = (a – ib) 20

∴ a 2 – b 2 =a and 2 ab = – 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.