Complex numbers. Operations

In this topic, we will take a look at the main operations on complex numbers: addition, multiplication, multiplication by scalar and division.

Addition of complex numbers

Suppose we have two complex numbers: $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$. Then their sum will be:
$z=z_1+z_2=(x_1+iy_1) + (x_2+iy_2)=(x_1+x_2)+i(y_1+y_2)$.

This means that in order to add one complex number to another we need to perform additions of their real and imaginary parts respectively. It goes without saying that subtraction of complex numbers is the same: we subtract real and imaginary parts of one number from another.

For example:

If $z_1=1+2i, z_2=2-i$, then $z=z_1-z_2 = (1+2i) - (2-i)=(1-2)+i(2-(-1))=-1+3i$.

Multiplication of complex numbers

Multiplication is also rather intuitive and easy. Assuming that $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, their product will be:

$z=z_1 \cdot z_2=(x_1+iy_1) \cdot (x_2+iy_2)=x_1 \cdot x_2-y_1 \cdot y_2+iy_1 \cdot x_2+iy_2 \cdot x_1$.

We simply open the brackets and multiply all elements in pairs. The sum of all elements without the imaginary part will make the real part of the complex number, while the sum of all elements with the imaginary part will contribute to the imaginary part of the result. However, it’s important to note that when we do $y_1 \cdot y_2$ we add a minus since $i \cdot i =-1$.

Consider the following example: $z_1=1+2i, z_2=2-i$, then $z=z_1 \cdot z_2 = (1+2i) \cdot (2-i)=2+2+4i-i=4+3i$.

Neutral element in multiplication is a number, multiplication by which will give the same number. For complex numbers it's $1+0i$ or simply $1$. It's clear that if we multiply any complex number by $1$, then it will stay the same.

Multiplication by scalar

This case is even simpler. Here we just multiply both real and imaginary parts by a scalar. So, for example, if we multiply $z=x+iy$ by a scalar $k$, we will get $kz=kx+iky$.

For instance: $z=5-4i, k=-7$. The answer is $kz=-35+28i$.

Division of complex numbers

Now division is where things get a bit more difficult. In order to understand how division is performed, we are going to introduce an additional equation. Earlier you've learned about conjugated numbers: the conjugate of $z = x + iy$ is $\overline{z} = x - iy$. Now let’s calculate $z \overline{z}$ using the common method: just by multiplying two complex numbers.

$z \overline{z}=(x+iy) \cdot (x-iy)=x^2+y^2+xyi-xyi=x^2+y^2$. So, the product of the number and its conjugated number is the sum of squares of its real and imaginary parts.

Time to get to the actual division! Say, we have a fraction $\frac {x_1+iy_1}{x_2+iy_2}$. Let’s multiply it by a fraction which is equal to $1$ but both numerator and denominator have the conjugated number to the initial denominator, which is $\frac{x_2-iy_2}{x_2-iy_2}$. This way, since we know the product of the number and its conjugated number, we know what the resulting denominator would be:

$\frac {z_1}{z_2}=\frac{x_1+iy_1}{x_2+iy_2}=\frac {(x_1+iy_1)(x_2-iy_2)}{(x_2+iy_2)(x_2-iy_2)}= \frac{x_1x_2+y_1y_2+i(y_1x_2-x_1y_2)}{x_2^2+y_2^2}=\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}+\frac{y_1x_2-x_1y_2}{x_2^2+y_2^2}i$

And for example: $\frac{2−i}{1+i}$:

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

Conclusion

In this topic, we've discussed the main arithmetic operations on complex numbers. This knowledge will help you during adding, multiplying and dividing complex numbers. Of course, these are not all existing operations, only the main ones. For instance, there are root or logarithm operations, but they are rather time-consuming, that is why they were omitted here. Although in case you’re interested you can read up about them on the internet.