Operations with Complex Numbers

All complex numbers have an imaginary part (the part with $i$ ) and a real part (the part which is a real number). Some operations with two complex numbers will result in another complex number while some may result in a real or imaginary number.

Addition and Subtraction of Complex Numbers

When adding or subtracting two complex numbers, you add/subtract the real and the imaginary parts seperately. If you treat it as though you are combining like terms, you will end up with the same answer in the end.

Example Find $(-5+3i) + (2-4i)$ .

Solution:

Since the operation outside of the parentheses is addition, it is OK to drop the parentheses. Doing this, you would have

$-5+3i+2-4i = -3-i$ by combining like terms. Therefore $(-5+3i) + (2-4i) = -3-i$

It is with subtraction that you must be more careful since negative signs distribute across parentheses.

Example Find $(2+9i)-(1-7i)$ .

Solution:

Distribute the negative! Afterwards, you can find the solution in the same way as the previous example was done.

$(2+9i)-(1-7i) = 2+9i -1 +7i = 1+16i$ .

Is it possible to add two complex numbers and “just” get a real number or imaginary number?

Certainly! The imaginary parts may cancel out or the real parts may cancel out. Remember, a real number can be thought of as an imaginary number with imaginary part zero. For instance, you could write $3$ as $3+0i$ . You could also write any imaginary number as a complex number with real part zero: $2i$ can be thought of as the same as $0+2i$ . Even though this is true, you will not really see it done.

Multiplication of Complex Numbers

Imagine that you wanted to find $(a+bi)(c+di)$ . Even if you didn’t know anything about complex numbers hopefully you would notice that these are two binomials and you have to use FOIL to multiply them! In fact, this is exactly how you multiply two complex numbers.

Example Find $(3-2i)(4-i)$

Solution:

Using FOIL will give the following unsimplified answer.

$(3-2i)(4-i) = (3)(4)+(3)(-i)+(-2i)(4)+(-2i)(-i) = 12-3i-8i+2i^2$ . By definition, $i^2=-1$ therefore this can be simplifed.

$12-3i-8i+2i^2 = 12 -3i-8i+2(-1) = 12-3i-8i-2=10-11i$ .

Example Find $(1+i)(-2+2i)$

Solution:

$(1+i)(-2+2i)$ $=(1)(-2)+(1)(2i)+(i)(-2)+(i)(2i)$ $=-2+2i-2i+2i^2$ $=-2+2i-2i-2$ $=-4$ .

Dividing Complex Numbers

The process for dividing complex numbers is very different from the methods discussed so far. It is very important to understand that you

DO NOT DIVIDE THE REAL AND IMAGINARY PARTS SEPARATELY!

Ok, ok – but then what do you do? We are going to borrow an idea from simplifying radicals and rationalize the denominator by multiplying the numerator and denominator by the denominator’s conjugate. Every complex number has a conjugate and the conjugate of $a+bi$ is $a-bi$ . This process will get rid of any imaginary terms in the denominator and allow you to give a final answer in the form $a+bi$ .