Question

Prove that all the complex number z which satisfy the equation z^n = (1 + z)^n (n>1) lie on a line parallel to imaginary axis.

Answer 1

Let $Z=\frac{1+z}{z}=1+\frac{1}{z}\ne 1$
$\therefore Z^n=\frac{(1+z{)}^2}{z^n}=1\left(given\right)orZ=(1{)}^{1/n}$
$cos\frac{2r\pi }{n}+isin\frac{2r\pi }{n}$
where r = 0, 1, 2, 3, .....(n - 1)
Now for r = 0, Z = 1 but since Z $\ne$ 1, hence we have
$\therefore Z=(cos\frac{2r\pi }{n}+isin\frac{2r\pi }{n})$
See Q.5 (d) and Q.6 P. 61-64.
or $1+\frac{1}{z}=cos\frac{2r\pi }{n}+isin\frac{2r\pi }{n}$
$\therefore \frac{1}{z}=-1+cos\frac{2r\pi }{n}isin\frac{2r\pi }{n}$
$=-2si{n}^2\frac{r\pi }{n}i2sin\frac{r\pi }{n}cos\frac{r\pi }{n}$
$=2isin\frac{r\pi }{n}(cos\frac{r\pi }{n}+isin\frac{r\pi }{n})$
as -1 = $i^2$
Taking reciprocal, we get
$\therefore z=\frac{1}{2isin\frac{r\pi }{n}}(cos\frac{r\pi }{n}-isin\frac{r\pi }{n})$
De-moivre's theorem
or $x+iy=-\frac{1}{2}-\frac{i}{2}cot\frac{r\pi }{n}\because \frac{1}{i}=-i$
$\therefore x=-\frac{1}{2}.$
This represents a line parallel to y-axis.