Forms of Complex Numbers | DP IB Maths: AA HL Revision Notes 2021

Modulus-Argument (Polar) Form

The Cartesian form of a complex number, $z = x + i y$ , is written in terms of its real part, $x$ , and its imaginary part, $y$
If we let $r = | z |$ and $θ = \arg z$ , then it is possible to write a complex number in terms of its modulus, $r$ , and its argument, $θ$ , called the modulus-argument (polar) form, given by...
- $z = r (\cos θ + isin θ)$
- This is often written as z = r cis θ
- This is given in the formula book under Modulus-argument (polar) form and exponential (Euler) form
It is usual to give arguments in the range $- π < θ \leq π$ or $0 \leq θ < 2 π$
- Negative arguments should be shown clearly
- e.g. $z = 2 (\cos (- \frac{π}{3}) + isin (- \frac{π}{3})) = 2 cis (- \frac{π}{3})$
  - without simplifying $\cos (- \frac{π}{3})$ to either $\cos (\frac{π}{3})$ or $\frac{1}{2}$
The complex conjugate of r cis θ is r cis (-θ )
If a complex number is given in the form $z = r (\cos θ - isin θ)$ , then it is not in modulus-argument (polar) form due to the minus sign
- It can be converted by considering transformations of trigonometric functions
  - $- \sin θ = \sin (- θ)$ and $\cos θ = \cos (- θ)$
- So $z = r (\cos θ - isin θ) = z = r (\cos (- θ) + isin (- θ)) = r cis (- θ)$
To convert from modulus-argument (polar) form back to Cartesian form, evaluate the real and imaginary parts
- E.g. $z = 2 (\cos (- \frac{π}{3}) + isin (- \frac{π}{3}))$ becomes $z = 2 (\frac{1}{2} + i (- \frac{\sqrt{3}}{2})) = 1 - \sqrt{3} i$

The main benefit of writing complex numbers in modulus-argument (polar) form is that they multiply and divide very easily
To multiply two complex numbers in modulus-argument (polar) form we multiply their moduli and add their arguments
- $|z_{1} z_{2}| = |z_{1}| |z_{2}|$
- $\arg (z_{1} z_{2}) = \arg z_{1} + \arg z_{2}$
So if z₁ = r₁ cis (θ₁) and z₂ = r₂ cis (θ₂)
- z₁ z₂= r₁r₂ cis (θ₁ + θ₂)
Sometimes the new argument, $θ_{1} + θ_{2}$ , does not lie in the range $- π < θ \leq π$ (or $0 \leq θ < 2 π$ if this is being used)
- An out-of-range argument can be adjusted by either adding or subtracting $2 π$
- E.g. If $θ_{1} = \frac{2 π}{3}$ and $θ_{2} = \frac{π}{2}$ then $θ_{1} + θ_{2} = \frac{7 π}{6}$
- This is currently not in the range $- π < θ \leq π$
- Subtracting $2 π$ from $\frac{7 π}{6}$ to give $- \frac{5 π}{6}$ , a new argument is formed
  - This lies in the correct range and represents the same angle on an Argand diagram
The rules of multiplying the moduli and adding the arguments can also be applied when…
- …multiplying three complex numbers together, $z_{1} z_{2} z_{3}$ , or more
- …finding powers of a complex number (e.g. $z^{2}$ can be written as $z z$ )
The rules for multiplication can be proved algebraically by multiplying z₁ = r₁ cis (θ₁) by z₂ = r₂ cis (θ₂), expanding the brackets and using compound angle formulae

To divide two complex numbers in modulus-argument (polar) form, we divide their moduli and subtract their arguments
- $|\frac{z_{1}}{z_{2}}| = \frac{|z_{1}|}{| z_{2} |}$
- $\arg (\frac{z_{1}}{z_{2}}) = \arg z_{1} - \arg z_{2}$
So if z₁ = r₁ cis (θ₁) and z₂ = r₂ cis (θ₂) then
- $\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} cis (θ_{1} - θ_{2})$
Sometimes the new argument, $θ_{1} - θ_{2}$ , can lie out of the range $- π < θ \leq π$ (or the range $0 < θ \leq 2 π$ if this is being used)
- You can add or subtract $2 π$ to bring out-of-range arguments back in range
The rules for division can be proved algebraically by dividing z₁ = r₁ cis (θ₁) by z₂ = r₂ cis (θ₂) using complex division and the compound angle formulae

Remember that r cis θ only refers to $r (\cos θ + isin θ)$
- If you see a complex number written in the form $z = r (\cos θ - isin θ)$ then you will need to convert it to the correct form first
- Make sure you are confident with basic trig identities to help you do this

Let $z_{1} = 4 \sqrt{2} cis \frac{3 π}{4}$ and $z_{2} = \sqrt{8} (\cos (\frac{π}{2}) - isin (\frac{π}{2}))$

Find

z_{1} z_{2}

, giving your answer in the form

r (\cos θ + isin θ)

where

0 \leq θ < 2 π

1-9-2-ib-aa-hl-forms-of-cn-we-solution-1-a

Find

\frac{z_{1}}{z_{2}}

, giving your answer in the form

r (\cos θ + isin θ)

where

- π \leq θ < π

1-9-2-ib-aa-hl-forms-of-cn-we-solution-1-b

This relates to the modulus-argument (polar) form as $z = r e^{i θ} = r cis θ$
This shows a clear link between exponential functions and trigonometric functions
This is given in the formula booklet under 'Modulus-argument (polar) form and exponential (Euler) form'

However in exponential form other arguments can be used and the same convention of adding or subtracting 2π can be applied

Euler's form allows for quick and easy multiplication and division of complex numbers
If $z_{1} = r_{1} e^{i θ_{1}}$ and $z_{2} = r_{2} e^{i θ_{2}}$ then
- $z_{1} \times z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$
  - Multiply the moduli and add the arguments
- $\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})}$
  - Divide the moduli and subtract the arguments
Using these rules makes multiplying and dividing more than two complex numbers much easier than in Cartesian form
When a complex number is written in Euler's form it is easy to raise that complex number to a power
- If $z = r e^{i θ}$ ,