User:J D Bowen/Math725 HW1

1. Let $$a, b\in \mathbb{R} \ $$. We endeavor to find the multiplicative inverse of the complex number $$z=a+bi\in\mathbb{C} \ $$. To do this, we assume that at least one of $$a, b \neq 0 \ $$; otherwise we have $$z=0 \ $$ and $$0^{-1} \ $$ is undefined.

We know matrix inverses are easy to compute, and we recall the matrix formulation of the complex numbers as

$$a+bi \cong \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix} \ $$.

This matrix has inverse

$$\frac{1}{a^2+b^2} \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} = \begin{pmatrix} \frac{a}{a^2+b^2} & \frac{b}{a^2+b^2} \\ \frac{-b}{a^2+b^2} & \frac{a}{a^2+b^2} \\ \end{pmatrix} \ $$,

which implies that the complex number defined as $$c+di, c= \frac{a}{a^2+b^2}, d= \frac{-b}{a^2+b^2} \ $$ is the multiplicative inverse of $$z=a+bi \ $$.

We check our suspicion:

$$(a+bi)(c+di) = (a+bi)(\frac{a}{a^2+b^2}-\frac{bi}{a^2+b^2}) = \frac{a^2-abi+abi+b^2}{a^2+b^2} = \frac{a^2+b^2}{a^2+b^2} = 1 \ $$, provided $$a^2+b^2 \neq 0 \ $$, which is guaranteed by our supposition that at least one of $$a,b \neq 0 \ $$.

2.