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Complex Numbers

Polar Form Terminology

The form in which we usually see complex numbers in is called rectangular form:

\(z=a+bi\),

where \(a,b\) are real numbers and \(i=\sqrt{-1}\).

However, we can also represent complex numbers in polar form:

\(z=r(cos\theta+isin\theta)\),

where \(r\) is called the modulus and \(\theta\) is called the argument of the complex number.

Sometimes, you might see \(z=r(cos\theta+isin\theta)\) abbreviated as \(z=rcis\theta\).

We will explain how to derive this form and convert between Rectangular and Polar form in the below sections.

Rectangular Form to Polar Form

Recall that we can represent complex numbers graphically on the complex plane (see Graphing Complex Numbers):

Notice that if we draw a line connect the origin \((0,0)\) to the point representing our complex number \((a,b)\), then draw another line from that point down to the Real axis, we can form a right-angled triangle:

                                                                                       

We can then label this magnitude/length of the line connecting the origin to the point \((a,b)\) as \(r\) (modulus) and the angle (argument) formed between that line and the Real axis as \(\theta\):

                                                                                          

Finally, we can use trigonometry to help us derive the formulas to calculate the modulus and argument.

Modulus

We can apply the Pythagorean Theorem to the right-angled triangle we identified above:

\(a^2+b^2=r^2\)

We can then rearrange this equation and isolate for \(r\), which gives us the formula to find the modulus of a complex number given its rectangular form (i.e. \(a+bi\)):

\(r=\sqrt{a^2+b^2}\)

*Note: Sometimes, you may see \(r\) equal to or expressed as \(|z|\) since the above formula is the same as the one for finding the magnitude of a complex number.

Tip: Since \(r\) represents the distance between two points, it must be a positive real number, so when calculating for this value, you can double check your work by making sure \(r \geq 0\)!

Argument

Recall the basic trigonometric ratio for tangent (you can further review this here):

\(tan(\theta)=\frac{opposite}{adjacent}\)

Applying these ratios to the above right-angled triangle gives:

\(tan(\theta)=\frac{b}{a}, \ a\ne 0\)

Then, isolating for \(\theta\) gives:

\(\theta=tan^{-1}(\frac{b}{a}), \ a\ne 0\)

If it's not specified, then you can choose to express the argument in radians or degrees.

*Note: Since the \(tan(\theta)\) function has a period of \(\pi\), you will encounter two possible values for \(\theta\). To determine which angle is the one you're looking for, identify the quadrant where the plotted point representing the complex number would be.

 

Example: Convert \(z=-3-3i\) to polar form.

See the below video for the solution!

 

Polar Form to Rectangular Form

To convert from polar form to rectangular form, we can use the right-angled triangle on the complex plane that we mentioned in the above section again.

                                                                                      

Recall the sine and cosine trigonometric ratios:

\(sin(\theta)=\frac{opposite}{hypotenuse}, \ \cos(\theta)=\frac{adjacent}{hypotenuse}\)

Applying these ratios to the right-angled triangle gives:

\(sin(\theta)=\frac{b}{r}, \ \cos(\theta)=\frac{a}{r}\)

Finally, we can isolate for \(a\) and \(b\) by multiplying both sides of each equation by \(r\):

\(a=rcos(\theta), \ b=rsin(\theta)\)

*Note: In addition to representing complex numbers with (Re, Im)-coordinates, using the two equations we derived above, we can also express them with polar coordinates: \((rcos(\theta), rsin(\theta))\).

 

Example: Convert \(z=3cis(\frac{\pi}{6})\) into rectangular form.

Solution:

1. We need to determine the modulus and argument of the complex number. In this case, we have:

\(r=3,\ \theta=\frac{\pi}{6}\)

2. Substitute \(r\) and \(\theta\) into the two formulas to solve for \(a\) and \(b\):

\(a=3cos(\frac{\pi}{6})\)

\(b=3sin(\frac{\pi}{6})\)

3. Evaluate the values of \(a\) and \(b\), then substitute them into the rectangular form:

\(a=3(\frac{\sqrt{3}}{2})=\frac{3\sqrt{3}}{2}\)

\(b=3(\frac{1}{2})=\frac{3}{2}\)

\(\therefore z=3cis(\frac{\pi}{6})\) in rectangular form is \(z=\frac{3\sqrt{3}}{2}+\frac{3}{2}i\).

Practice

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