# Understanding quotients of complex numbers

I am reading an old complex variables textbook which states:

Given $z = a + bi$, $z_1 = a_1 + b_1i$, and $z_2 = a_z + b_2i \neq 0$, we have $z = \dfrac{z_1}{z_2} = \dfrac{a_1a_2 + b_1b_2}{a_2^2 + b_2^2} + i \dfrac{b_1a_2 - a_1 b_2}{a_2^2 + b_2^2}$.

I understand how $a_1 = a_2a-b_2b$ and $b_1 = b_2a + a_2 b$ however,

I don't understand how $\dfrac{z_1}{z_2} = \dfrac{a_1a_2 + b_1b_2}{a_2^2 + b_2^2} + i \dfrac{b_1a_2 - a_1 b_2}{a_2^2 + b_2^2}$.

The authors also mention that $a_2^2 + b_2^2$ is a determinant that is non-zero however I'm missing the significance of this fact.

Any help is appreciated. Thanks in advance.

• If you have confirmed that $a_1 = a_2a-b_2b$ and $b_1 = b_2a + a_2 b$, then you know why $\frac{z_1}{z_2} = \frac{a_1a_2 + b_1b_2}{a_2^2 + b_2^2} + i \frac{b_1a_2 - a_1 b_2}{a_2^2 + b_2^2}$: because $z=\frac{z_1}{z_2}$ is precisely the number such that $z_2z = z_1.$ How someone might have found that expression for $\frac{z_1}{z_2}$ is shown in the answers. – David K May 23 '15 at 2:48

$$\frac{z_1}{z_2}=\frac{a_1+b_1i}{a_2+b_2i}\cdot\frac{a_2-b_2i}{a_2-b_2i}=\frac{(a_1+b_1i)(a_2-b_2i)}{a_2^2+b_2^2}$$
$$\frac{a_1+ib_1}{a_2+ib_2} = \frac{a_1+ib_1}{a_2+ib_2}\frac{a_2-ib_2}{a_2-ib_2} = \frac{(a_1+ib_1)(a_2-ib_2)}{(a_2+ib_2)(a_2-ib_2)} = \frac{(a_1a_2+b_1b_2)+i(a_2b_1-a_1b_2)}{a_2^2+b_2^2} .$$
The part about $a_2^2+b_2^2$ being a "determinant" refers to the fact that a complex number $a+bi$ can be represented as the real $2\times 2$ matrix $$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$ In this representation, addition and multiplication of complex numbers correspond exactly to matrix addition and matrix multiplication! (And additionally the first column of the matrix is just the vector that represents the complex number in the complex plane.)
The determinant of the matrix turns out to be $a^2+b^2$ which is $0$ if and only if the complex number is $0$.