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2.06 Complex numbers


What is a complex number?

The history of complex numbers begins, as with many stories in mathematics, with the ancient Greeks. They were the first to try and find solutions to polynomials, and they discovered that none of the numbers they knew of could solve ones like $x^2+1=0$x2+1=0. These sorts of polynomials arose from seemingly solvable, physical problems - a famous one was proposed by Diophantus of Alexandria (AD 210 – 294 approx):

“A right triangle has a perimeter of $12$12 units and an area of $7$7 squared units. What are the lengths of its sides?”

Here is such a triangle:

Letting $AB=x$AB=x and $AC=h$AC=h, the area of the triangle is then given by

$\text{Area }=\frac{1}{2}xh$Area =12xh

and the perimeter is 

$\text{Perimeter }=x+h+\sqrt{x^2+h^2}$Perimeter =x+h+x2+h2

Try this yourself before revealing the solution:

Using the information provided above, show that the equations for perimeter and area can be reduced to this polynomial:


Does this polynomial have real solutions? What does that mean for Diophantus' triangle?

Another mathematician named Jerome Cardan (1501 - 1576) also thought about problems like these.  He tried to solve the problem of finding two numbers, $a$a and $b$b, whose sum is $10$10 and whose product is $40$40:


Eliminating $b$b gives $a\left(10-a\right)=40$a(10a)=40, which distributes to $a^2-10a+40=0$a210a+40=0. Solving this quadratic would give the solutions:


But since there are no "real" numbers whose square is $-15$15, the term $\sqrt{-15}$15 has no meaning in the real number system - we say there are no real solutions to the problem Cardan posed. That said, if we treat the "numbers" $a=5+\sqrt{-15}$a=5+15 and $b=5-\sqrt{-15}$b=515 like ordinary numbers, they are solutions! They sum to $10$10:


... and they multiply to $40$40:


In each instance the $\sqrt{-15}$15 term disappeared, canceling nicely as we performed the algebra. So everything in mathematics still works, and most mathematicians of the time thought this was a silly, slightly eerie "trick". It wasn’t until the nineteenth century that the power of these sorts of solutions began to be fully understood. 

The advent of the complex numbers

While complex numbers were created almost by accident when solving a series of abstract questions, they have ended up being critical to a broad range of applications in contemporary mathematics.  Wi-fi systems, telephone networks, electrical circuits, electromagnetism, any areas that use physics and differential equations together and more all rely heavily on complex numbers.  

The other key area is in any field that relies on the theory of self-similarity, commonly referred to as fractals. It is a very modern field, requiring enormous computing power to even represent, and the results are stunningly beautiful. Computer-generated imagery in movies and games have complex numbers as a foundational cornerstone. 

This fractal is the famous Mandelbrot Set.


Complex numbers are built on the concept that there is an object, called $i$i, that is the square root of $-1$1.

$i=\sqrt{-1}\equiv i^2=-1$i=1i2=1

This is not a so-called "real" number, since the square of any real number is always positive. But using this single object we can define an entire new dimension for the number line.

In the previous section, we came across the value $5+\sqrt{-15}$5+15. Rewriting this as $5+\sqrt{-1\times15}=5+\sqrt{-1}\times\sqrt{15}$5+1×15=5+1×15 then allows us to express it more concisely:


Instead of inventing a new symbol for "the square root of $-15$15", we just re-use the symbol $i$i from before. Generally, all complex numbers $z$z can be written in the form $z=x+iy$z=x+iy, where $x$x and $y$y are real (and therefore familiar) numbers. And just like $$ is used to denote the set of all real numbers, we use the symbol $$ to denote all the complex numbers.

So a number like $5+3i$5+3i is a complex number.  It has both real and imaginary components. 


The real part of $z$z is $Re\left(z\right)=x$Re(z)=x, and the imaginary part of $z$z is $Im\left(z\right)=y$Im(z)=y.

So for $5+3i$5+3i$Re\left(5+3i\right)=5$Re(5+3i)=5 and $Im\left(5+3i\right)=3$Im(5+3i)=3.

Every real number $x$x can be written as $x+i0$x+i0, which means every real number is also a complex number - in other words, the set of real numbers is a subset of the set of complex numbers.  

Writing numbers in complex form

To be able to write some numbers in complex form, a little algebraic manipulation may be necessary, mostly involving the fact that $\sqrt{-1}=i$1=i or that $-1=i^2$1=i2.

Worked example

Question 1

Rewrite  $\sqrt{-50}$50 in the form $z=x+iy$z=x+iy

$\sqrt{-50}$50 $=$= $\sqrt{-1\times50}$1×50

using properties of radicals:
$\sqrt{ab}=\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$ab=a×b=a×b

  $=$= $\sqrt{-1}\times\sqrt{50}$1×50  
  $=$= $i\sqrt{50}$i50

and now simplify the $\sqrt{50}$50:

  $=$= $i5\sqrt{2}$i52  

So in the form $x+iy$x+iy, the real component is $x=0$x=0 and the imaginary component is $y=5\sqrt{2}$y=52, so $x+iy=0+i5\sqrt{2}$x+iy=0+i52.

Practice questions


Consider the complex number $8+7i$8+7i.

  1. What is the real part?

  2. What is the imaginary part?

  3. Identify the type(s) of number this is.



    nonreal complex


    pure imaginary



Write down the complex number that has a real part $0$0 and an imaginary part $\sqrt{5}$5.


Express $\sqrt{-80}$80 in terms of $i$i.


Operations with negative radicals

Properties of Radical Numbers 

Recall that a square root sign is actually an index.  It is the power half. 


Remember the properties we have already seen for our four basic operations with square roots.  

Multiplication and Division with Square Roots

Multiplication of Square Roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ab=a×b

Division of Square Roots $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ab=ab

Now, what about addition and subtraction I hear you ask.  Well, there are no simplification laws for general addition and subtraction of square roots, unless the value in the radicand is the same.

Addition and Subtraction of Square Roots

Addition and subtraction only works if the value in the square root is identical. 






But  $\sqrt{a}+\sqrt{b}$a+b is definitely NOT equal to $\sqrt{a+b}$a+b


When the square root value is negative

Combining together now our ability to convert $i=\sqrt{-1}$i=1, and our knowledge of the properties of radicals we can simplify a whole myriad of questions involving negative radicals. 

Things to remember when doing this

  • Separate a value like $-49$49 in under the square root sign into two components, $-1\times49$1×49 and then take the square root of each component
  • Always simplify the radical as far as you can, look for square numbers that could be factors
  • Be careful when simplifying the product of two negative radicals. 

Here is an example of what we mean by that last point

Worked example

Question 5

Simplify $\sqrt{-2}\times\sqrt{-30}$2×30

$\sqrt{-2}\times\sqrt{-30}$2×30 $=$= $\sqrt{-1\times2}\times\sqrt{-1\times30}$1×2×1×30
$=$= $i\sqrt{2}\times i\sqrt{30}$i2×i30
$=$= $i^2\sqrt{2\times30}$i22×30
$=$= $-\sqrt{4\times15}$4×15
$=$= $-2\sqrt{15}$215
Watch out!

A very common error for students to make here is to assume that the negative 2 multiplied by the negative 30 would give positive 60 and hence they would end up with the answer of $2\sqrt{15}$215.  So deal with each $\sqrt{-1}$1 component separately.  

Practice questions


Express $\sqrt{-100}$100 in terms of $i$i.


Express $-\sqrt{-29}$29 in terms of $i$i.


Find the value of $\frac{\sqrt{-33}\times\sqrt{-3}}{\sqrt{11}}$33×311.


Addition and subtraction of complex numbers

Complex numbers can be added and subtracted very easily, following the normal laws of algebra. 

We must only add and subtract like terms.

And in complex numbers, like terms are the real parts and the imaginary parts.


Let's look a an example with two complex numbers $z_1=3+2i$z1=3+2i and $z_2=6-4i$z2=64i

To add these two complex numbers together we must first identify the real and imaginary components of each. 


Then we add the real components, and imaginary components respectfully


Worked examples

Question 9
$\left(3+7i\right)+\left(2+i\right)$(3+7i)+(2+i) $=$= $\left(2+3\right)+\left(7+1\right)i$(2+3)+(7+1)i
  $=$= $5+8i$5+8i

See how in this example I grouped the real and imaginary parts and then added.  Sometimes this helps keep track of all the components.  Sometimes you can jump straight to the answer. 


Question 10
$\left(9-2i\right)-\left(-2+6i\right)$(92i)(2+6i) $=$= $9-2i+2-6i$92i+26i
  $=$= $11-8i$118i

In this example, I distributed the parentheses observing the change of sign and then collected like terms. 

Practice questions


Evaluate $\left(3+6i\right)+\left(7+3i\right)$(3+6i)+(7+3i).


Evaluate $\left(6+9i\right)-\left(-3-4i\right)$(6+9i)(34i).


Evaluate $\left(-6\sqrt{7}-2i\right)+\left(3\sqrt{7}+7i\right)$(672i)+(37+7i).


Multiplication of complex numbers

When multiplying complex numbers, we apply algebraic conventions as well as the fact that $i^2=-1$i2=1.

We can multiply two single terms together

  • Real and Imaginary terms $/to$/to $3\times7i=21i$3×7i=21i
  • Two imaginary terms $/to$/to $6i\times4i=24i^2=-24$6i×4i=24i2=24 (remember that $i^2=-1$i2=1)


We can a term with a complex number containing both real and imaginary parts

  • Real number multiplied by a complex number $5\left(6+2i\right)=5\times\left(6+2i\right)=30+10i$5(6+2i)=5×(6+2i)=30+10i (just like distributing through parenthesis in algebra)
  • Complex term multiplied by a complex number  $3i\left(-4+6i\right)=-12i+18i^2=-12i-18=-18-12i$3i(4+6i)=12i+18i2=12i18=1812i


Worked example

Question 14

Use the distributive property to distribute and simplify $\left(1-3i\right)\left(4+2i\right)$(13i)(4+2i).

$\left(1-3i\right)\left(4+2i\right)$(13i)(4+2i) $=$= $\left(1\right)\left(4\right)+\left(-3i\right)\left(4\right)+\left(1\right)\left(2i\right)+\left(-3i\right)\left(2i\right)$(1)(4)+(3i)(4)+(1)(2i)+(3i)(2i)
  $=$= $4-12i+2i-6i^2$412i+2i6i2
  $=$= $4-10i-6i^2$410i6i2
  $=$= $4-10i-6\left(-1\right)$410i6(1)
  $=$= $4-10i+6$410i+6
  $=$= $10-10i$1010i


Question 15

Use the distributive property to distribute and simplify $\left(4-3i\right)\left(4+3i\right)$(43i)(4+3i).

$\left(4-3i\right)\left(4+3i\right)$(43i)(4+3i) $=$= $16-9i^2$169i2
  $=$= $16+9$16+9
  $=$= $25$25

This final case is a special example as the numbers are what we call conjugates of each other.  We will study more about these special cases later. 


Try this yourself before checking out the solution

We already know the radical laws such as 



Extend these laws to see what happens if $a$a or $b$b or both $a$a&$b$b are <$0$0

(see here for the solution)

Practice questions


Simplify $-6\left(3-5i\right)$6(35i).


Simplify $\sqrt{10}i\left(8+\sqrt{10}i\right)$10i(8+10i), writing your answer in terms of $i$i.


Simplify $\left(2+5i\right)\left(5i-2\right)$(2+5i)(5i2).


Simplify $-2i\left(4-3i\right)^2$2i(43i)2, leaving your answer in terms of $i$i.


Powers of i


Before we embark on further explorations with our complex number $i$i, let's have a look at what happens when we take successive powers of $i$i.

Try this activity yourself first, before checking out my solution.

Make a list of powers of $i$i, up to $i^{15}$i15.

Simplify the results. 

Then generalize the pattern. 

(see here for the solution)


Now that you have explored how to simplify powers of $i$i, the only other thing to do is combine this with other algebraic simplifications. 


Worked examples

Question 20

Simplify $(2i)^4$(2i)4

We need to remember our exponent laws here, so $\left(ab\right)^n=a^n\times b^n$(ab)n=an×bn


$\left(2\times1\right)^4$(2×1)4 $=$= $2^4\times i^4$24×i4
  $=$= $16\times i^2\times i^2$16×i2×i2
  $=$= $16\times\left(-1\right)\times\left(-1\right)$16×(1)×(1)
  $=$= $16$16


Question 21

Simplify $4i^2-3i^3-7i^4$4i23i37i4

$4i^2-3i^3-7i^4$4i23i37i4 $=$= $4\times-1-3\times-i-7\times1$4×13×i7×1
  $=$= $-4--3i-7$43i7
  $=$= $3i-11$3i11


Practice questions


Simplify $2i^7$2i7.


Simplify $\left(2i\right)^9$(2i)9.


Simplify $\left(\sqrt{5}i\right)^6$(5i)6.


Complex conjugates


A complex conjugate is defined as a number with equal magnitudes of real and imaginary parts, but opposite in sign.  


Let's have a look at some conjugate pairings

$6+\frac{1}{4}i$6+14i $6-\frac{1}{4}i$614i
$-15-3i$153i $-15+3i$15+3i
$1-i$1i $1+i$1+i
$26i$26i $-26i$26i
$7$7 $7$7


We have dealt with a similar concept before when we were studying radicals.  We underwent a process called rationalizing the denominator.  

Just like this process, we use complex conjugates to help us define a process for division of complex numbers.  

Instead of a formal division process, we use conjugates to turn the operation of division into an operation of multiplication. 

Let's look at this question as an example. $\frac{2+3i}{4+5i}$2+3i4+5i



So, to tackle the division we 

  • multiply by a fraction that is equivalent to 1
  • use the conjugate of the denominator to construct this fraction
  • multiply the binomials on the numerator and denominators and simplify
  • the denominator will always simplify to $a^2-b^2$a2b2 (for the distribution $\left(a+bi\right)\left(a-bi\right)$(a+bi)(abi))


Let's finish off this question $\frac{2+3i}{4+5i}$2+3i4+5i

$\frac{2+3i}{4+5i}$2+3i4+5i $=$= $\frac{2+3i}{4+5i}\times\frac{4-5i}{4-5i}$2+3i4+5i×45i45i
  $=$= $\frac{\left(2+3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}$(2+3i)(45i)(4+5i)(45i)
  $=$= $\frac{8+12i-10i-15i^2}{16+20i-20i-25i^2}$8+12i10i15i216+20i20i25i2
  $=$= $\frac{8+2i-15\left(-1\right)}{16-25\left(-1\right)}$8+2i15(1)1625(1)
  $=$= $\frac{8+2i+15}{16+25}$8+2i+1516+25
  $=$= $\frac{23+2i}{41}$23+2i41
  $=$= $\frac{23}{41}+\frac{2i}{41}$2341+2i41

This last step may or may not be necessary.  Sometimes it's easier to do further work when the real and imaginary parts are separate.  


Worked examples

Question 25
$\frac{4}{i}$4i $=$= $\frac{4}{i}\times\frac{-i}{-i}$4i×ii
  $=$= $\frac{-4i}{-i^2}$4ii2
  $=$= $\frac{-4i}{1}$4i1
  $=$= $-4i$4i


Question 26
$\frac{1-i}{1+i}$1i1+i $=$= $\frac{1-i}{1+i}\times\frac{1-i}{1-i}$1i1+i×1i1i
  $=$= $\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$(1i)(1i)(1+i)(1i)
  $=$= $\frac{1-2i+i^2}{1+1}$12i+i21+1
  $=$= $\frac{1-2i-1}{2}$12i12
  $=$= $\frac{-2i}{2}$2i2
  $=$= $-i$i


Question 27
$\frac{3}{6+i}+\frac{2}{3-2i}$36+i+232i $=$= $\frac{3}{6+i}\times\frac{6-i}{6-i}+\frac{2}{3-2i}\times\frac{3+2i}{3+2i}$36+i×6i6i+232i×3+2i3+2i
  $=$= $\frac{3\left(6-i\right)}{37}+\frac{2\left(3+2i\right)}{13}$3(6i)37+2(3+2i)13
  $=$= $\frac{18-3i}{37}+\frac{6+4i}{13}$183i37+6+4i13
  $=$= $\frac{456+109i}{481}$456+109i481


Practice questions


Find the value of $\frac{4+6i}{1+i}$4+6i1+i.


Find the value of $\frac{4+7i}{2+i}$4+7i2+i.


Find the value of $\frac{2-3i}{2+3i}$23i2+3i.



Understand there is a complex number i such that i^2 = -1, and every complex number has form a + bi where a and b are real numbers.


Use the relation i^2 = -1 and the commutative, associative, and distributive properties to add, subtract and multiply complex numbers.

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