If n is a positive integer greater than unity | vedclass

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Question

(A) Given the equation $z^n = (z + 1)^n$,we can write $\left( \frac{z}{z + 1} \right)^n = 1$.This implies that $\frac{z}{z + 1}$ is an $n$-th root of

Vedclass · Accepted Answer

$	ext{Re}(z) < 0$

Vedclass · Answer

(A) Given the equation $z^n = (z + 1)^n$,we can write $\left( \frac{z}{z + 1} ight)^n = 1$.This implies that $\frac{z}{z + 1}$ is an $n$-th root of unity.Taking the modulus on both sides,we get $\left| \frac{z}{z + 1} ight| = 1$,which means $|z| = |z + 1|$.Let $z = x + iy$. Then $|x + iy| = |x + 1 + iy|$.Squaring both sides,we get $x^2 + y^2 = (x + 1)^2 + y^2$.$x^2 = x^2 + 2x + 1$.$2x + 1 = 0$,which gives $x = -1/2$.Since $x = 	ext{Re}(z) = -1/2$,it follows that $	ext{Re}(z) < 0$.

If $n$ is a positive integer greater than unity and $z$ is a complex number satisfying the equation $z^n = (z + 1)^n$,then

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