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:
$6x^243x+84=0$6x2−43x+84=0
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:
$a+b=10$a+b=10, $ab=40$ab=40
Eliminating $b$b gives $a\left(10a\right)=40$a(10−a)=40, which distributes to $a^210a+40=0$a2−10a+40=0. Solving this quadratic would give the solutions:
$a=\frac{1}{2}\left(10\pm\sqrt{60}\right)=5\pm\sqrt{15}$a=12(10±√60)=5±√−15
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=5−√−15 like ordinary numbers, they are solutions! They sum to $10$10:
$a+b=\left(5+\sqrt{15}\right)+\left(5\sqrt{15}\right)=\left(5+5\right)+\left(\sqrt{15}\sqrt{15}\right)=10+0=10$a+b=(5+√−15)+(5−√−15)=(5+5)+(√−15−√−15)=10+0=10
... and they multiply to $40$40:
$ab=\left(5+\sqrt{15}\right)\left(5\sqrt{15}\right)=5^2+5\times\sqrt{15}5\times\sqrt{15}\left(\sqrt{15}\right)^2=25\left(15\right)=40$ab=(5+√−15)(5−√−15)=52+5×√−15−5×√−15−(√−15)2=25−(−15)=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.
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. Wifi 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 selfsimilarity, commonly referred to as fractals. It is a very modern field, requiring enormous computing power to even represent, and the results are stunningly beautiful. Computergenerated 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=√−1≡i2=−1
This is not a 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:
$5+\sqrt{15}=5+i\sqrt{15}$5+√−15=5+i√15.
Instead of inventing a new symbol for "the square root of $15$−15", we just reuse 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.
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.
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: 
$=$=  $\sqrt{1}\times\sqrt{50}$√−1×√50  
$=$=  $i\sqrt{50}$i√50 
and now simplify the $\sqrt{50}$√50: 

$=$=  $i5\sqrt{2}$i5√2 
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=5√2, so $x+iy=0+i5\sqrt{2}$x+iy=0+i5√2.
Consider the complex number $8+7i$8+7i.
What is the real part?
What is the imaginary part?
Identify the type(s) of number this is.
real
nonreal complex
pure imaginary
real
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.
Recall that a square root sign is actually an index. It is the power half.
$\sqrt{a}=a^{\frac{1}{2}}$√a=a12
Remember the properties we have already seen for our four basic operations 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=√a√b
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 only works if the value in the square root is identical.
Thus
$a\sqrt{c}+b\sqrt{c}=\left(a+b\right)\sqrt{c}$a√c+b√c=(a+b)√c
and
$a\sqrt{c}b\sqrt{c}=\left(ab\right)\sqrt{c}$a√c−b√c=(a−b)√c
But $\sqrt{a}+\sqrt{b}$√a+√b is definitely NOT equal to $\sqrt{a+b}$√a+b
Combining together now our ability to convert $i^2=\sqrt{1}$i2=√−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
Here is an example of what we mean by that last point
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 
$=$=  $i\sqrt{2}\times i\sqrt{30}$i√2×i√30  
$=$=  $i^2\sqrt{2\times30}$i2√2×30  
$=$=  $\sqrt{4\times15}$−√4×15  
$=$=  $2\sqrt{15}$−2√15 
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}$2√15. So deal with each $\sqrt{1}$√−1 component separately.
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×√−3√11.
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 $z_2=64i$z2=6−4i
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
$\left(3+7i\right)+\left(2+i\right)$(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.
$\left(92i\right)\left(2+6i\right)$(9−2i)−(−2+6i)  $=$=  $92i+26i$9−2i+2−6i 
$=$=  $118i$11−8i 
In this example, I distributed the parentheses observing the change of sign and then collected like terms.
Evaluate $\left(3+6i\right)+\left(7+3i\right)$(3+6i)+(7+3i).
Evaluate $\left(6+9i\right)\left(34i\right)$(6+9i)−(−3−4i).
Evaluate $\left(6\sqrt{7}2i\right)+\left(3\sqrt{7}+7i\right)$(−6√7−2i)+(3√7+7i).
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
We can a term with a complex number containing both real and imaginary parts
Use the distributive property to distribute and simplify $\left(13i\right)\left(4+2i\right)$(1−3i)(4+2i).
$\left(13i\right)\left(4+2i\right)$(1−3i)(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) 
$=$=  $412i+2i6i^2$4−12i+2i−6i2  
$=$=  $410i6i^2$4−10i−6i2  
$=$=  $410i6\left(1\right)$4−10i−6(−1)  
$=$=  $410i+6$4−10i+6  
$=$=  $1010i$10−10i 
Use the distributive property to distribute and simplify $\left(43i\right)\left(4+3i\right)$(4−3i)(4+3i).
$\left(43i\right)\left(4+3i\right)$(4−3i)(4+3i)  $=$=  $169i^2$16−9i2 
$=$=  $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
$\sqrt{a}\times\sqrt{b}=\sqrt{ab}$√a×√b=√ab
$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$√a√b=√ab
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)
Simplify $6\left(35i\right)$−6(3−5i).
Simplify $\sqrt{10}i\left(8+\sqrt{10}i\right)$√10i(8+√10i), writing your answer in terms of $i$i.
Simplify $\left(2+5i\right)\left(5i2\right)$(2+5i)(5i−2).
Simplify $2i\left(43i\right)^2$−2i(4−3i)2, leaving your answer in terms of $i$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.
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
Thus,
$\left(2\times1\right)^4$(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 
Simplify $4i^23i^37i^4$4i2−3i3−7i4
$4i^23i^37i^4$4i2−3i3−7i4  $=$=  $4\times13\timesi7\times1$4×−1−3×−i−7×1 
$=$=  $43i7$−4−−3i−7  
$=$=  $3i11$3i−11 
Simplify $2i^7$2i7.
Simplify $\left(2i\right)^9$(2i)9.
Simplify $\left(\sqrt{5}i\right)^6$(√5i)6.
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$6−14i 
$153i$−15−3i  $15+3i$−15+3i 
$1i$1−i  $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
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{45i}{45i}$2+3i4+5i×4−5i4−5i 
$=$=  $\frac{\left(2+3i\right)\left(45i\right)}{\left(4+5i\right)\left(45i\right)}$(2+3i)(4−5i)(4+5i)(4−5i)  
$=$=  $\frac{8+12i10i15i^2}{16+20i20i25i^2}$8+12i−10i−15i216+20i−20i−25i2  
$=$=  $\frac{8+2i15\left(1\right)}{1625\left(1\right)}$8+2i−15(−1)16−25(−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.
$\frac{4}{i}$4i  $=$=  $\frac{4}{i}\times\frac{i}{i}$4i×−i−i 
$=$=  $\frac{4i}{i^2}$−4i−i2  
$=$=  $\frac{4i}{1}$−4i1  
$=$=  $4i$−4i 
$\frac{1i}{1+i}$1−i1+i  $=$=  $\frac{1i}{1+i}\times\frac{1i}{1i}$1−i1+i×1−i1−i 
$=$=  $\frac{\left(1i\right)\left(1i\right)}{\left(1+i\right)\left(1i\right)}$(1−i)(1−i)(1+i)(1−i)  
$=$=  $\frac{12i+i^2}{1+1}$1−2i+i21+1  
$=$=  $\frac{12i1}{2}$1−2i−12  
$=$=  $\frac{2i}{2}$−2i2  
$=$=  $i$−i 
$\frac{3}{6+i}+\frac{2}{32i}$36+i+23−2i  $=$=  $\frac{3}{6+i}\times\frac{6i}{6i}+\frac{2}{32i}\times\frac{3+2i}{3+2i}$36+i×6−i6−i+23−2i×3+2i3+2i 
$=$=  $\frac{3\left(6i\right)}{37}+\frac{2\left(3+2i\right)}{13}$3(6−i)37+2(3+2i)13  
$=$=  $\frac{183i}{37}+\frac{6+4i}{13}$18−3i37+6+4i13  
$=$=  $\frac{456+109i}{481}$456+109i481 
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{23i}{2+3i}$2−3i2+3i.
Know there is a complex number I such that i^2 = 1, and every complex number has the form a + bi with a and b real.
Use the relation i^2 = 1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.