Multiplicative Inverses of Complex Numbers

We have seen how to add, subtract, and multiply complex numbers, but it remains to learn how to divide them. The key is to think of division by a number $z$ as multiplying by the multiplicative inverse of $z$. You are probably already familiar with this concept for ordinary real numbers: dividing by $2$ is the same as multiplying by $\frac12$, dividing by 3 is the same as multiplying by $\frac13$, and so on. Algebraically, we write $\frac{x}{k} = x \cdot \frac1k$.

For complex numbers, the multiplicative inverse can be deduced using the complex conjugate. We have already seen that multiplying a complex number $z=a+bi$ with its complex conjugate $\overline{z}=a-bi$ gives the real number $a^2+b^2$. 

Thus, we have $z\overline{z}=a^2+b^2$, and dividing through by $a^2+b^2$ gives:

$z \cdot \frac{\overline{z}}{a^2+b^2}=1$ 

So the multiplicative inverse of $z$ must be the complex conjugate of $z$ divided by its modulus squared. We can write $\frac{1}{z}= \frac{\overline{z}}{\abs{z}^2}$

Example 1

The multiplicative inverse of $1+2i$ is:

$\frac{1-2i}{1+4}=(1/5)-(2/5)i$ 

To see that this is correct, we can multiply these numbers to see if we get the multiplicative identity number $1$. Using FOIL, we have:

 $\begin {aligned}(1+2i)((1/5)-(2/5)i) &= 1/5 -(2/5)i+(2/5)i+4/5 \\&=1/5+4/5 \\&=1 \end {aligned}$

Example 2

The multiplicative inverse of $3-4i$ is:

$\frac{3+4i}{9+16} = (3/25)+(4/25)i$ 

Again, checking through multiplication, we have:

 $\begin {aligned}(3-4i)((3/25)+(4/25)i&=9/25 -(12/25)i+(12/25)i+16/25 \\&=\frac{25}{25} \\&=1 \end {aligned}$

Division of Complex Numbers

Suppose you wanted to divide the complex number $z=2+3i$ by the number $1+2i$. 

Since dividing by $1+2i$ is the same as multiplying by the multiplicative inverse (which we have seen above is $(1/5)-(2/5)i$), we have:

 $\frac{2+3i}{1+2i}=(2+3i)((1/5)-(2/5)i)$

If we multiply out (FOIL) this last expression we obtain:

$2/5 -(4/5)i+(3/5)i+(6/5)=(8/5)-(1/5)i$ 

In general, for complex numbers $w$ and $z$, we have $\frac{w}{z} =w \cdot \frac{1}{z} = w\cdot \frac{\overline{z}}{\abs{z}^2}$

As an example, let's use this formula directly to compute $\frac{1-i}{3-4i}$ 

 $\begin {aligned}\frac{1-i}{3-4i} &= (1-i)\cdot\frac{3+4i}{9+16} \\&=(1-i)((3/25)+(4/25)i) \end {aligned}$

This multiplies out to be $\begin {aligned}(1-i)((3/25)+(4/25)i) &= 3/25 +(4/25)i-(3/25)i+4/25 \\&= 7/25 + (1/25)i \end {aligned}$