First slide

Algebra of complex numbers

Question

If $z = x + iy (x, y \in R, x \neq - 1 / 2)$ the number of values of z satisfying $| z |^{n} = z^{2} | z |^{n - 2} + z | z |^{n - 2} + 1 \cdot (n \in N, n > 1)$ is

Difficult

A

0
B

1
C

2
D

3

Solution

The given equation is
$\begin{array}{l} | z |^{n} = (z^{2} + z) | z |^{n - 2} + 1 \\ \Rightarrow z^{2} + z is real \\ \Rightarrow z^{2} + z = {\bar{z}}^{2} + \bar{z} \\ \Rightarrow (z - \bar{z}) (z + \bar{z} + 1) = 0 \\ \Rightarrow z = \bar{z} = x as z + \bar{z} + 1 \neq 0 (x \neq - 1 / 2) \end{array}$
Hence, the given equation reduces to
$\begin{array}{l} x^{n} = x^{n} + x | x |^{n - 2} + 1 \\ \Rightarrow x | x |^{n - 2} = - 1 \\ \Rightarrow x = - 1 \end{array}$
So, the number of solutions is 1.

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