Finding an expression for the complex number Z^-1

Question

So I want to find out an expression to express: $$z^{-1}$$ I know the answer is: $$z^{-1} = \frac{x-iy}{x^2+y^2}$$

But how would I go about proving this/the steps to this?

Answer 1

Hint: $$\frac{1}{z}=\frac{\bar{z}}{z\bar{z}}.$$

Answer 2

Our intuition from the real numbers tells us that the inverse of a real number $c$ is $1/c$.

Likewise, we conjecture that the inverse of a complex $z \in \mathbb{C}$ is $\frac{1}{z} = \frac{1}{x + iy}$. To get this into a more recognizable form, multiply the numerator and denominator by $z^* = x - iy$. From here, we arrive at $\frac{x-iy}{x^2+y^2} = \frac{x}{x^2 + y^2} + \frac{y}{x^2+y^2}i$. Certainly, our purported $z^{-1}$ is of the form $a+bi$ for $a, b \in \mathbb{R}$. Next, we can simply multiply $z$ by our purported $z^{-1}$ to confirm that we do indeed get $1 + 0i$, which is the multiplicative identity in $\mathbb{C}$.

Answer 3

Set $z = x + iy$; then

$\dfrac{1}{z} = \dfrac{1}{ x + iy}. \tag{1}$

Multiply by

$1 = \dfrac{x - iy}{x - iy}: \tag{2}$

$\dfrac{1}{z} = \dfrac{(x - iy) \times 1}{(x - iy)(x + iy)} = \dfrac{x - iy}{x^2 + y^2}, \tag{3}$

since $(x - iy)(x + iy) = x^2 + y^2$. And that does it!

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!