complex numbers Notes

By the end of this chapter you should be familiar with:

  • The properties of complex numbers
  • Cartesian form of complex numbers
  • Polar form of complex numbers
  • Exponential form of complex numbers
  • Powers and roots of complex numbers
  • The use of complex numbers in STEM applications

Real numbers are simply the combination of rational and irrational numbers, in the number system. Complex numbers cannot be represented on the number line, but are analytic solutions to equations whose solutions are not real numbers. These complex numbers have an imaginary unit.


A number that is expressed in terms of the square root of a negative number is called an imaginary number. The imaginary unit is i and i = √(-1)


  • i = √(-1)
  • i2 = -1
  • i3 = -i
  • i4 = 1, then i5 = i and so on for in

Every power n is an integer multiple of four that produces in = 1 and the pattern repeats.

Example: Express each in its simplest form
a) i32 = i4 = 1
b) i126 = i2 = -1

Complex number has two parts, a real part and an imaginary part which can be thought of as two- dimensional numbers. This concept can be understood by placing them in the complex plane called the argand plane. A complex number has the form z = a + bi.

|z| is the modulus of the complex number and the argument of z is defined as the angle between the positive real axis and the line from origin to z.

Example:Find the zeros of the equation y = x2 – 2x + 5 and find the modulus and argument of the zeros.

Solution: Arg(x) = tan-1(2/1) = 63.435o

When a quadratic polynomial in x has zeros r1 and r2 then (x – r1) and (x – r2) are its factors. Then the quadratic equation is = x2 – (r1 + r2)x + r1r2 Therefore, we can conclude that the negative sum of the zeroes is the coefficient of x and the product of the zeros is the constant term.

The complex conjugate of z = a + bi is z* = a – bi and the product of zz* is (a + bi)(a – bi) = a2 + b2

Example: Rationalize in the form a + bi.

Solution: 6/(1+𝑖) = (6/(1+𝑖)) × ((1−𝑖)/(1− 𝑖)) = 6(1−𝑖)/2
= 3(1 – i) = 3 – 3i


The polar form of a complex number z = a + bi is z = r(cosθ + isinθ) and can be abbreviated as z = r cis  

where r=|z|= √(𝒂𝟐 + 𝒃𝟐)(modulus) a = rcosθ and b = rsinθ
θ = tan-1(b/a) (argument also referred to as arg(z)) for a > 0 and = tan-1(b/a) + πor θ = tan-1(b/a) + 180° for a < 0. Example: Express 5 + 2i complex number in polar form.

Solution: The polar form of a complex number z = a + bi is z=r(cosθ + isinθ).
So, first find the absolute value of r.
r=|z|= √(𝒂𝟐 + 𝒃𝟐) = √(5𝟐 + 2𝟐) = √29 = 5.39
Now find the argument θ.
Since a > 0, use the formula θ=tan-1(b/a).
θ = tan-1(2/5) = 0.38

Note that here θ is measured in radians.

Therefore, the polar form of 5 + 2i is about 5.39(cos(0.38) + isin(0.38)).


Euler’s formula is the statement that e= cos(θ) + isin(θ). When x = π, we get Euler’s identity, eiπ = -1, or eiπ + 1 = 0. Basically, z = a + bi = reExample: Taking the last example where we found r = 5.39 and = 0.38.5 + 2i can also be expressed in Euler’s form as z = 5.39 ei0.38


DeMoivre’s Theorem states that zn = (re)n = rne
or zn = rn cis(nθ)

Example: Compute (3 + 3i)5

r=|z|= √(𝒂𝟐 + 𝒃𝟐) = √(3𝟐 + 3𝟐) = 3√2 = tan-1(3/3) = π/4
zn = (re)n
(3 + 3i)5 = (3√2)5 𝑒𝑖 5𝜋/4 = 972√2(cos(5π/4) + i sin(5π/4)) = -972 – 972i



Using the complex plane, we can represent voltages across resistors, capacitors and inductors.

The voltage across the resistor is regarded as a real quantity, while the voltage across an inductor is regarded as a positive imaginary quantity, and across a capacitor we have a negative imaginary quantity. Our axes are as follows:All AC waveforms have sinusoidal curves.


The impedance of a circuit is the total effective resistance to the flow of current by a combination of the elements of the circuit.

To find this total voltage, we cannot just add the voltages VR, VL and VC.

Because VL and VC are considered to be imaginary quantities, we have:

Impedance VRLC = IZ

Z = R + j(X­L ​– XC​)

|Z| =√𝑹𝟐 + (𝑿𝑳 − 𝑿𝑪)𝟐

tan θ = (𝑿𝑳 − 𝑿𝑪)/𝑹

Example: A circuit has a resistance of 5Ω in series with a reactance across an inductor of 3Ω. Represent the impedance by a complex number, in polar form.

Solution: In this case, XL​= 3Ω and XC ​= 0 so XL ​– XC​ = 3 Ω.
So in rectangular form, the impedance is written: Z=5+3j Ω

Using calculator, the magnitude of Z is given by: 5.83, and the angle θ is given by: 30.96∘.

So, the voltage leads the current by 30.96∘, as shown in the diagram.
Presenting Z as a complex number (in polar form), we have: Z=5.83∠30.96∘ Ω.