Cartesian/polar forms | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

\begin{aligned} z & = x + i y = r (cos θ + i sin θ) \\ = r cis θ \end{aligned}

Cartesian form gives coordinates. Polar form gives modulus $r$ and argument $θ$ .

Two descriptions of one point

On an Argand diagram, $x$ is the horizontal real coordinate and $y$ is the vertical imaginary coordinate. The modulus $r$ is the distance from the origin.

The argument $θ$ is measured anticlockwise from the positive real axis. Here angles are in radians. The abbreviation $cis θ$ means $cos θ + i sin θ$ .

x = r cos θ, y = r sin θ

Equal-scale Argand plane with point 3+3i, horizontal and vertical components 3, modulus 3 square root 2, and argument pi divided by four.

Locate the point before choosing its angle

Use Pythagoras for the modulus. For a nonzero real coordinate, $tan θ = \frac{y}{x}$ , but a tangent ratio alone cannot distinguish opposite quadrants.

Find a reference angle, then use the signs of $x$ and $y$ to select the correct quadrant.

r = \sqrt{x^{2} + y^{2}}

Keep the direction meaningful

Angles differing by a whole turn give the same point. State an argument in the interval requested by the question.

The number zero has no defined argument. Its modulus is zero, so every direction reaches the same point.

cis (θ + 2 k π) = cis θ (k \in ℤ)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Convert a second-quadrant point

Write $- \sqrt{3} + i$ in polar form.

Worked solution

1
Find the modulus.
$r = \sqrt{{(- \sqrt{3})}^{2} + 1^{2}} = 2$
The modulus is a nonnegative distance.
2
Locate the quadrant and reference angle.
$x < 0, y > 0, α = \frac{π}{6}$
The point lies in quadrant II. Its reference triangle has side lengths $\sqrt{3}$ , $1$ and $2$ .
3
Choose the argument and write polar form.
$\begin{aligned} θ & = π - \frac{π}{6} = \frac{5 π}{6} \\ z & = 2 cis (\frac{5 π}{6}) \end{aligned}$
The angle is measured from the positive real axis, not from the negative one.

Answer

$2 cis (\frac{5 π}{6})$ .

EXAMPLE 02

Recover exact coordinates

Write $4 cis (- \frac{π}{3})$ in Cartesian form.

Worked solution

1
Calculate the real coordinate.
$x = 4 cos (- \frac{π}{3}) = 2$
Cosine gives the horizontal component.
2
Calculate the imaginary coordinate.
$y = 4 sin (- \frac{π}{3}) = - 2 \sqrt{3}$
The negative angle lies below the real axis, so this coordinate is negative.
3
Combine the coordinates.
$z = x + i y = 2 - 2 \sqrt{3} i$
Keep $i$ with the imaginary component, not with the modulus.

Answer

$2 - 2 \sqrt{3} i$ .

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Read an argument from an axis

Give a polar form of $- 3 i$ .

Show answer & reasoning

$3 cis (- \frac{π}{2})$ .

The point is $3$ units below the origin: modulus $3$ and argument $- \frac{π}{2}$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Convert $1 + i$ to polar form.
2 marks
Convert $- 2 - 2 i$ to polar form with $- π < θ \leq π$ .
2 marks
Convert $6 cis (\frac{π}{6})$ to Cartesian form.
2 marks
A signal is represented by $5 cis π$ . State its real and imaginary components.
2 marks
A student writes $- 1 + i = \sqrt{2} cis (- \frac{π}{4})$ . Explain and correct the error.
3 marks
Why is an argument of $z = 0$ undefined?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Convert $1 + i$ to polar form.
2 marks
$r = \sqrt{1 + 1} = \sqrt{2}$ and the equal positive coordinates give $θ = \frac{π}{4}$ .
$\sqrt{2} cis (\frac{π}{4})$ .
Convert $- 2 - 2 i$ to polar form with $- π < θ \leq π$ .
2 marks
The modulus is $\sqrt{8} = 2 \sqrt{2}$ . Quadrant III has principal angle $- \frac{3 π}{4}$ .
$2 \sqrt{2} cis (- \frac{3 π}{4})$ .
Convert $6 cis (\frac{π}{6})$ to Cartesian form.
2 marks
Multiply 6 by $cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ and $sin (\frac{π}{6}) = \frac{1}{2}$ .
$3 \sqrt{3} + 3 i$ .
A signal is represented by $5 cis π$ . State its real and imaginary components.
2 marks
$5 cos π = - 5$ and $5 sin π = 0$ .
Real component $- 5$ ; imaginary component $0$ .
A student writes $- 1 + i = \sqrt{2} cis (- \frac{π}{4})$ . Explain and correct the error.
3 marks
The given angle produces a positive real part and a negative imaginary part, opposite to those required.
The angle is in the wrong quadrant; use $\sqrt{2} cis (\frac{3 π}{4})$ .
Why is an argument of $z = 0$ undefined?
2 marks
Every value of $θ$ gives $0 cis θ = 0$ , so no direction is distinguished.
Zero has no direction from the origin.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the method.

Two descriptions of one point

Locate the point before choosing its angle

Keep the direction meaningful

See the method in action.

Convert a second-quadrant point

Find the modulus.

Locate the quadrant and reference angle.

Choose the argument and write polar form.

Recover exact coordinates

Calculate the real coordinate.

Calculate the imaginary coordinate.

Combine the coordinates.

Try it yourself.

Read an argument from an axis

Keep practising