We are given a complex number $z = x + iy$, where $z$ is non-zero. We need to: a) Express $\frac{1}{z}$ in the form $a+ib$. b) Given that $z + \frac{1}{z} = k$, where $k$ is a real number, prove that either $z$ is real or $|z| = 1$.

1. Problem Description

We are given a complex number $z = x + iy$, where $z$ is non-zero. We need to:

a) Express $\frac{1}{z}$ in the form $a+ib$.

b) Given that $z + \frac{1}{z} = k$, where $k$ is a real number, prove that either $z$ is real or $|z| = 1$.

2. Solution Steps

a) To express $\frac{1}{z}$ in the form $a+ib$, we have:

$\frac{1}{z} = \frac{1}{x+iy}$

To get rid of the complex number in the denominator, multiply both the numerator and denominator by the conjugate of the denominator:

$\frac{1}{z} = \frac{1}{x+iy} \cdot \frac{x-iy}{x-iy} = \frac{x-iy}{x^2 + y^2} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$

So, $a = \frac{x}{x^2+y^2}$ and $b = \frac{-y}{x^2+y^2}$.

b) Given $z + \frac{1}{z} = k$, where $k$ is a real number. We know that $z = x + iy$.

Then $z + \frac{1}{z} = x + iy + \frac{1}{x+iy} = x + iy + \frac{x-iy}{x^2+y^2} = k$

$x + iy + \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2} = k$

$(x + \frac{x}{x^2+y^2}) + i(y - \frac{y}{x^2+y^2}) = k$

Since $k$ is a real number, the imaginary part must be zero.

$y - \frac{y}{x^2+y^2} = 0$

$y(1 - \frac{1}{x^2+y^2}) = 0$

This gives two cases:

Case 1: $y=0$. If $y=0$, then $z = x$, which means $z$ is real.

Case 2: $1 - \frac{1}{x^2+y^2} = 0$

$\frac{1}{x^2+y^2} = 1$

$x^2 + y^2 = 1$

Since $|z| = \sqrt{x^2+y^2}$, we have $|z| = \sqrt{1} = 1$.

Therefore, either $z$ is real or $|z|=1$.

3. Final Answer

a) $\frac{1}{z} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$

b) Either $z$ is real or $|z|=1$.