© 2025 Perplex Learning Incorporated. All rights reserved.
All content on this website has been developed independently from and is not endorsed by the International Baccalaureate Organization. International Baccalaureate and IB are registered trademarks owned by the International Baccalaureate Organization.
In the previous sections, we've used tools such as rationalizing complex fractions to write complex numbers in Cartesian form, and in doing so have translated complex numbers into a number and graphing system we already recognize. But we can also write complex numbers in entirely new forms that make more sense for working with these values. In this section, we'll introduce some important formulas and use trigonometry to represent complex numbers in different ways.
First, let's define some more terms.
Discussion
Remember that we can think of the complex modulus ∣z∣=√a2+b2 as the length of the hypotenuse of a right triangle constructed by plotting z on the complex plane. Consider the value
Powered by Desmos
How could we find the value of θ in the triangle above?
We label the horizontal leg a and the vertical leg b. From the right triangle
so
We call the value of θ the argument of a complex number z.
Complex Argument
The argument of a complex number is the angle that it forms with the real (x) axis on the complex plane:
Powered by Desmos
By noticing a right angled triangle, we can say that
When a>0:
Note on convention: By convention, the argument is usually given in radians in the range [−π,π]. It will be made clear by the IB which range is preferred in a given question.
Example
Find the argument of 1+i.
Because arctan(θ) is only defined for −π≤θ≤π, if z is in the second or third quadrant, we cannot calculate the argument using arctan(θ). However there is clearly still an angle between z and the origin in this case, so we use the fact that the angle from the y-axis to z forms an alternate interior angle with one in the range of arctan to calculate arg(z).
Complex argument when a<0
If a<0, then argz is in the second or third quadrant, which are not in the range of arctan. We therefore need to add or subtract π to get the correct argument:
Powered by Desmos
When a<0:
When z is in the second quadrant, we add π; when z is in the third quadrant, we subtract π.
Note on convention: By convention, the argument is usually given in radians in the range [−π,π]. It will be made clear by the IB which range is preferred in a given question.
Example: Find the argument of z=−√3−i.
Since b<0, z is in the third quadrant, and we will subtract π.
Checkpoint
Find the argument θ∈(−π,π] of the complex number −3−√3i.
Select the correct option
When z is purely imaginary (a=0), ab is undefined. However, there is clearly still an angle made between the x-axis and z. We thereby define the argument as either 2π or −2π.
Complex Argument when a=0 (purely imaginary)
If a=0, then tan(argz)=ab is undefined. tanθ is also undefined for θ=2π,23π… So when a=0 we have
This can be seen on the complex diagram by remembering that bi lies on the yi axis:
Powered by Desmos
Note on convention: By convention, the argument is usually given in radians in the range [−π,π]. It will be made clear by the IB which range is preferred in a given question.
Example
Find the argument of z=−3i.
We have z=0−3i, so z is purely imaginary and lies along the y-axis. Since −3<0, the argument of z is argz=−2π.
Checkpoint
It is given that z=2i
Find the value of arg(z).
Select the correct option
Discussion
The diagram below shows how the point z changes as you shift the values of ∣z∣ and arg(z).
Powered by Desmos
If we know the values of ∣z∣ and arg(z), can we find the coordinates of z?
Any point z in the plane sits at the end of a segment from the origin of length ∣z∣, making an angle arg(z) with the positive real axis. If you drop a perpendicular from z down to the real axis, you form a right-angled triangle whose – horizontal side is the “run” x, – vertical side is the “rise” y, – hypotenuse is the distance ∣z∣.
By basic trigonometry in that triangle,
So the point z has coordinates
Polar form of complex numbers
The modulus ∣z∣ and argument argz uniquely define the complex number z. That means we can represent any complex number using its modulus and argument instead of a+bi:
Powered by Desmos
It is conventional to call r=∣z∣ and θ=argz. Using trigonometry, we deduce that
And we use the shorthand cisθ=cosθ+isinθ:
Example
Express z=2−2i in the form rcisθ.
First we find r=∣z∣=√22+(−2)2=2√2.
Next, we find θ=argz=arctan(−22)=−4π.
Thus
Exercise
Find 2−2√3i in the form rcis(θ).
Select the correct option
Discussion
Now that we know there is a way to describe complex numbers in terms of r and θ, you might wonder if there's a simpler function than (cosθ+isinθ) that describes the angle θ -- some kind of "direction function," call it D(θ).
If that function existed, how would we use it to describe any complex number z?
Any non-zero complex number can be written as its magnitude times a unit “direction.” If we let
and choose θ so that D(θ) is the unit complex number pointing in the direction of z, then
because D(θ) plays the role of cosθ+isinθ.
This function does exist. In particular, it turns out that
so we can actually write any complex number in the form
We call this Euler's form, or sometimes exponential form. After you learn calculus, you will be able to prove that reiθ and rcisθ are equivalent.
Euler's form
There is one more way to express complex numbers:
For example, we can write z=−1=1⋅cis(π)=eiπ. This leads to the classic result
We call this Euler's form (or sometimes exponential form) because of the presence of Euler's number, e.
Example
Find z=3eiπ/6 in cartesian form.
We have
Exercise
It is given that z=43eiπ/6
Find z in the form z=a+bi for a,b∈R.
Select the correct option
To see why this form is useful, let's try an example problem that gives numbers in Euler's form.
Discussion
Let z=keiπ/2,w=3eiπ.
Find the value of k that makes the following equation true:
(Hint: Remember to use exponent rules!)
We have
Then
We require
so comparing moduli gives
Thus
As we just saw, using Euler's form makes multiplying and dividing complex numbers much easier, which is the main advantage of this form.
Multiplying in mod-arg form
The main advantage of Euler's form is that it makes multiplying complex numbers much easier:
In words, when we multiply two complex numbers the arguments add and the moduli multiply.
Similarly for division:
In words, when we divide one complex number from another, we subtract the arguments and divide the moduli.
Example
Given that z⋅(1+i)=(2−2i), find z.
First convert everything to Euler's form:
Checkpoint
It is given that z⋅(3eiπ/6)=eiπ/3
Find z in the form z=reiθ.
Select the correct option
Exercise
Let z=e−iπ/6 and w=3cis(2π).
Find zw∗ in the form a+bi.
Select the correct option
Discussion
Remember that when we graph a complex number z=a+bi on the complex plane, the conjugate z∗=a−bi is the reflection of z over the x-axis:
Powered by Desmos
Knowing this, what do you think the conjugate of z would be when written in the form
rcis(θ)?
reiθ?
When you reflect a point in the complex plane across the real (x-) axis, its distance from the origin stays the same but its “height” above the axis changes sign. Let’s see how that flips the angle.
A point
has Cartesian coordinates (rcosθ,rsinθ). Reflecting over the x-axis sends
so here we get
But cos(−θ)=cosθ and sin(−θ)=−sinθ, so
and hence the reflected point is
Intuitively, an angle θ measured counter-clockwise above the x-axis becomes an angle of the same size measured clockwise below the axis—that is, −θ.
In the exponential form z=reiθ, the same change of sign in the angle gives
Complex conjugate in polar form
If z=rcisθ=reiθ, then the conjugate z∗ is
Checkpoint
It is given that z=Γcis(2π).
Find the value of z∗.
Select the correct option
0 / 6
Being able to convert complex numbers from one form to another allows us to rewrite weird, unfriendly-looking trigonometric functions in familiar forms. In fact complex numbers have many practical applications wherever sinusoidal waves are found.
In particular, converting between mod-arg and Euler form allows us to add trigonometric functions with the same frequency but different phase.
Discussion
Consider the functions f(x)=3sin(x), g(x)=5sin(x+2π). Let h(x)=f(x)+g(x).
Our goal is to write h(x) in the form h(x)=Asin(x+B). Let's walk through how to do this step by step.
First, show that Im(rcis(θ)) is a sine function.
Now, use your knowledge of mod-arg and Euler forms to show that the function Asin(x+B) can be written in the form Im(AexieBi).
Do the same for f(x) and g(x) above, writing them in the form Im(AexieBi) (for the relevant values of A and B).
Use your Euler forms of f(x) and g(x) to write h(x) in the same form.
Finally, rewrite h(x) as a sine function, h(x)=Asin(x+B).
Part (a)
We use the definition
so
Taking the imaginary part gives
which is exactly a sine function of θ with amplitude r.
Part (b)
We begin with Euler’s formula
so that
Taking imaginary parts gives
But
hence
This is exactly the desired form.
Part (c)
From part (b) we have
For f(x)=3sinx we set A=3, B=0, so
For g(x)=5sin(x+2π) we set A=5, B=2π, so
Part (d)
Hence in the form Im(AeixeiB) we have AeiB=3+5eiπ/2, so
Part (e)
From part (d) we have
Write 3+5i in polar form: its modulus is
and its argument is
Hence
and
We can use a similar process to the one above to add any trigonometric functions with different phases.
Adding complex waves (phasors)
For any two complex waves f(x)=r1sin(ax+α1) and g(x)=r2sin(ax+α2) with the same frequency but different phase, we can use Euler's form of complex numbers to express h(x) as a single sine function:
So we can write
where
Note: In most questions about adding complex waves, you can use your graphing calculator to find the phase shift instead of deriving it by hand. An example of how this works is in the problem below.
!!
0 / 5
Chat
0 / 6
!!
0 / 5