Pre-Calculus Complex numbers (advanced)

Polar form of complex numbers

Q: What is the polar form of a complex number?

z equals r times cosine theta plus i sine theta, where r is the modulus and theta is the argument.

Q: How do you find the modulus and argument?

Modulus is the square root of a squared plus b squared; argument is arctangent of b over a, adjusted for the quadrant.

Q: Why adjust the argument for the quadrant?

Because arctangent returns only values in two quadrants; the actual point may lie elsewhere, so you add or subtract pi as needed.

Q: How do you convert polar back to rectangular?

Use a equals r cosine theta and b equals r sine theta, then write a plus b i.

20 practice questions 0 video lessons Theory + worked examples

Polar Form of Complex Numbers

California Pre-Calculus • Standard N-CN.4 • Complex Numbers

Polar Form of Complex Numbers is a topic in Complex Numbers in the California Common Core State Standards. It is aligned to Standard N-CN.4, which requires students to represent complex numbers in both rectangular and polar form.

The polar form of a complex number is \(r(\cos\theta+i\sin\theta)\), using its modulus \(r\) and argument \(\theta\).

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Practice questions

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Theory

Besides rectangular form \(a+bi\), a complex number can be written in polar (trigonometric) form using its distance and direction from the origin:

\[z=r(\cos\theta+i\sin\theta),\]

where \(r=|z|=\sqrt{a^2+b^2}\) is the modulus and \(\theta\) the argument (the angle from the positive real axis). This is often abbreviated \(z=r\,\text{cis}\,\theta\).

Mind the quadrant. \(\arctan\dfrac{b}{a}\) only gives the reference angle; adjust by the quadrant of \((a,b)\) to get the true argument.

Polar form uses the modulus \(r\) and argument \(\theta\).

Converting to polar form.

Conversions between the two forms:

\[r=\sqrt{a^2+b^2},\quad \theta=\arctan\dfrac{b}{a}\ (+\text{quadrant});\quad a=r\cos\theta,\ b=r\sin\theta\]

modulus is the square root of a squared plus b squared; argument is arctangent of b over a adjusted for quadrant

Polar form makes multiplication easy — the next topic multiplies moduli and adds arguments.

How to convert to polar form

Compute \(r\) from \(\sqrt{a^2+b^2}\).
Find the reference angle \(\arctan\left|\dfrac{b}{a}\right|\).
Adjust for the quadrant of \((a,b)\) to get \(\theta\).
Write \(z=r(\cos\theta+i\sin\theta)\).

Example 1 — Rectangular to polar

Write \(z=1+i\) in polar form.

Solution

Find the modulus and argument.

\(r\)	\(=\)	\(\sqrt{1^2+1^2}=\sqrt2\)
\(\theta\)	\(=\)	\(\arctan\dfrac{1}{1}=\dfrac{\pi}{4}\)
\(z\)	\(=\)	\(\sqrt2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)\)

polar form is root 2 cis pi over 4

Example 2 — Polar to rectangular

Write \(z=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)\) in rectangular form.

Solution

Evaluate the sine and cosine.

\(z\)	\(=\)	\(2\left(\dfrac12+i\dfrac{\sqrt3}{2}\right)\)
	\(=\)	\(1+\sqrt3\,i\)

rectangular form is 1 plus root 3 i

Example 3 — Argument in Quadrant II

Find the argument of \(z=-1+i\).

Solution

\(z\) is in Quadrant II, so add \(\pi\) to the reference angle.

\(\text{ref}\)	\(=\)	\(\arctan\dfrac{1}{1}=\dfrac{\pi}{4}\)
\(\theta\)	\(=\)	\(\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}\)

argument is three pi over four

Example 4 — Modulus and argument

Give the modulus and argument of \(z=3i\).

Solution

\(z=0+3i\) lies straight up the imaginary axis.

\(r\)	\(=\)	\(3\)
\(\theta\)	\(=\)	\(\dfrac{\pi}{2}\)

modulus 3, argument pi over 2

Common pitfalls

Adjust the argument for the quadrant. \(\arctan\) alone can put the angle in the wrong half of the plane.

\(r\) is nonnegative. Direction is carried entirely by \(\theta\).

Keep angles exact for special values like \(\dfrac{\pi}{4}\).

Frequently asked questions

What is the polar form of a complex number?

\(z=r(\cos\theta+i\sin\theta)\), where \(r\) is the modulus (distance from the origin) and \(\theta\) is the argument (angle).

How do you find the modulus and argument?

Modulus \(r=\sqrt{a^2+b^2}\); argument \(\theta=\arctan\dfrac{b}{a}\), adjusted for the quadrant of the point.

Why adjust the argument for the quadrant?

Because \(\arctan\) returns only values in two quadrants; the actual point may lie elsewhere, so you add or subtract \(\pi\) as needed.

How do you convert polar back to rectangular?

Use \(a=r\cos\theta\) and \(b=r\sin\theta\), then write \(a+bi\).

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Operations on complex numbers in polar form

Polar form of complex numbers

Polar Form of Complex Numbers

📖 Theory

How to convert to polar form

Common pitfalls

Frequently asked questions

Theory