This note shows two more ways to write a complex number: polar form, which uses a distance and an angle, and Euler form, which packs the same information into an exponential. Using them, multiplying and dividing complex numbers becomes much simpler than in Cartesian form.
Key ideas
On the Argand diagram, the point for \(z=a+bi\) can be described in two ways. You can give its Cartesian coordinates \((a,b)\). Or you can give its distance from the origin and the direction of the line joining it to the origin.
- The modulus \(r=|z|=\sqrt{a^2+b^2}\) is the distance from the origin.
- The argument \(\theta=\arg z\) is the angle the line makes with the positive real axis. Anticlockwise is positive and clockwise is negative.
From the right-angled triangle, \(a=r\cos\theta\) and \(b=r\sin\theta\). So \(z=r(\cos\theta+i\sin\theta)\). This is the modulus-argument form, also called polar form. It is often shortened to \(r\,\mathrm{cis}\,\theta\).
Euler form. The same number can be written \(z=re^{i\theta}\). This comes from Euler's formula \(e^{i\theta}=\cos\theta+i\sin\theta\). In these forms \(\theta\) must be in radians unless a question says otherwise. A special case is \(e^{i\pi}=-1\), which gives \(e^{i\pi}+1=0\).
Principal argument. Any angle \(\theta+2\pi k\) points in the same direction, so the argument is not unique. The principal argument is the one in the interval \(-\pi\lt\theta\le\pi\). Use it unless told otherwise.
Finding the argument. Do not just press \(\tan^{-1}(b/a)\) and stop. The calculator gives an angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), which is only correct when \(a\gt0\). Sketch the point first to see which quadrant it is in. Find the acute reference angle \(\tan^{-1}\left|\frac ba\right|\), then adjust for the quadrant.
- Quadrant 1: \(\theta=\) reference angle.
- Quadrant 2: \(\theta=\pi-\) reference angle.
- Quadrant 3: \(\theta=-\pi+\) reference angle.
- Quadrant 4: \(\theta=-\) reference angle.
Multiplying and dividing. If \(z_1=r_1e^{i\theta_1}\) and \(z_2=r_2e^{i\theta_2}\), then multiplying multiplies the moduli and adds the arguments. Dividing divides the moduli and subtracts the arguments. If the result has an argument outside \(-\pi\lt\theta\le\pi\), add or subtract \(2\pi\) to bring it back.
Why multiplying is simpler in polar form
In Cartesian form, \((a+bi)(c+di)\) needs four products and a collection of terms. In Euler form, the rules of indices do the work: \(re^{i\alpha}\times se^{i\beta}=rs\,e^{i(\alpha+\beta)}\). Geometrically, multiplying by \(se^{i\beta}\) stretches the point away from the origin by a factor \(s\) and rotates it anticlockwise by the angle \(\beta\). Multiplying by \(i=e^{i\pi/2}\) is therefore a quarter turn, and multiplying by \(-1=e^{i\pi}\) is a half turn.
Dividing undoes this. You shrink by the modulus of the divisor and rotate back by its argument. The conjugate of \(re^{i\theta}\) is \(re^{-i\theta}\), which is the reflection in the real axis, so \(zz^*=r^2\).
Exact values worth knowing
Many questions use angles with exact trigonometric values. Keep this short list in mind.
- \(\cos\frac\pi6=\frac{\sqrt3}2\), \(\sin\frac\pi6=\frac12\).
- \(\cos\frac\pi4=\sin\frac\pi4=\frac{\sqrt2}2\).
- \(\cos\frac\pi3=\frac12\), \(\sin\frac\pi3=\frac{\sqrt3}2\).
- The real numbers \(1\) and \(-1\) have arguments \(0\) and \(\pi\). The numbers \(i\) and \(-i\) have arguments \(\frac\pi2\) and \(-\frac\pi2\).
When converting between forms, check that your answer has the right size and direction. A number such as \(-2+2\sqrt3\,i\) has a negative real part and a positive imaginary part, so it must be in the second quadrant, which agrees with an argument of \(\frac{2\pi}3\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(r=|z|=\sqrt{a^2+b^2}\) | Modulus of \(z=a+bi\). |
| \(\theta=\arg z\), with \(-\pi\lt\theta\le\pi\) | Principal argument, measured from the positive real axis. |
| \(z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta\) | Modulus-argument (polar) form. |
| \(z=re^{i\theta}\) | Euler form. Needs \(\theta\) in radians. |
| \(a=r\cos\theta,\quad b=r\sin\theta\) | Converting from polar to Cartesian form. |
| \(z_1z_2=r_1r_2\,e^{i(\theta_1+\theta_2)}\) | Product: multiply moduli, add arguments. |
| \(\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}\) | Quotient: divide moduli, subtract arguments. |
Worked examples
Example 1 Write \(z=1-i\sqrt3\) in modulus-argument form and in Euler form.
The point \((1,-\sqrt3)\) is in the fourth quadrant.
Modulus: \(r=\sqrt{1^2+(\sqrt3)^2}=\sqrt4=2\).
Reference angle: \(\tan^{-1}\left(\dfrac{\sqrt3}{1}\right)=\dfrac\pi3\).
In the fourth quadrant the argument is negative, so \(\theta=-\dfrac\pi3\).
Answer: \(2\left(\cos\left(-\tfrac\pi3\right)+i\sin\left(-\tfrac\pi3\right)\right)=2e^{-i\pi/3}\)
Example 2 Write \(z=4\,\mathrm{cis}\left(\dfrac{2\pi}{3}\right)\) in the form \(a+bi\).
\(a=4\cos\dfrac{2\pi}3=4\times\left(-\dfrac12\right)=-2\).
\(b=4\sin\dfrac{2\pi}3=4\times\dfrac{\sqrt3}2=2\sqrt3\).
Answer: \(z=-2+2\sqrt3\,i\)
Example 3 Let \(z_1=2e^{3\pi i/4}\) and \(z_2=5e^{2\pi i/3}\). Find \(z_1z_2\) and \(\dfrac{z_1}{z_2}\), giving the arguments in the interval \(-\pi\lt\theta\le\pi\).
Product: moduli \(2\times5=10\). Arguments \(\dfrac{3\pi}4+\dfrac{2\pi}3=\dfrac{9\pi}{12}+\dfrac{8\pi}{12}=\dfrac{17\pi}{12}\).
This is greater than \(\pi\), so subtract \(2\pi\): \(\dfrac{17\pi}{12}-\dfrac{24\pi}{12}=-\dfrac{7\pi}{12}\).
Quotient: moduli \(\dfrac25=0.4\). Arguments \(\dfrac{9\pi}{12}-\dfrac{8\pi}{12}=\dfrac{\pi}{12}\), which is already in range.
Answer: \(z_1z_2=10e^{-7\pi i/12}\) and \(\dfrac{z_1}{z_2}=0.4\,e^{\pi i/12}\)
Common errors
- Using \(\tan^{-1}(b/a)\) without checking the quadrant. For \(-1+i\) the calculator gives \(-\frac\pi4\), but the correct argument is \(\frac{3\pi}{4}\).
- Working in degrees inside \(e^{i\theta}\). Euler form needs radians.
- Adding the moduli when multiplying, or multiplying the arguments. Moduli multiply, arguments add.
- Leaving an argument outside the principal range, such as \(\frac{17\pi}{12}\), when the question asks for the principal argument.
- Putting a minus sign in front of \(\cos\) or \(\sin\) in polar form, such as \(2(-\cos\theta+i\sin\theta)\). Polar form is \(r(\cos\theta+i\sin\theta)\) with \(r\gt0\); a negative sign belongs in the angle.
Exam tip: Sketch the Argand diagram before finding an argument. A thirty-second sketch shows the quadrant, so you can tell at once whether to add or subtract from the reference angle.
Check yourself
- State the principal argument of \(-5\).
Answer
The point \((-5,0)\) lies on the negative real axis, so \(\arg(-5)=\pi\). - Write \(-2-2i\) in the form \(re^{i\theta}\), with \(\theta\) the principal argument.
Answer
\(r=\sqrt{4+4}=2\sqrt2\). The point is in the third quadrant and the reference angle is \(\tan^{-1}1=\frac\pi4\), so \(\theta=-\pi+\frac\pi4=-\frac{3\pi}4\). Result: \(2\sqrt2\,e^{-3\pi i/4}\). - Write \(6e^{i\pi/6}\) in the form \(a+bi\).
Answer
\(a=6\cos\frac\pi6=6\times\frac{\sqrt3}2=3\sqrt3\) and \(b=6\sin\frac\pi6=3\). Result: \(3\sqrt3+3i\). - Given \(z=\sqrt2\,\mathrm{cis}\frac\pi4\) and \(w=3\,\mathrm{cis}\frac{\pi}{2}\), find \(zw\) in polar form and in the form \(a+bi\).
Answer
Multiply moduli: \(3\sqrt2\). Add arguments: \(\frac{3\pi}4\). So \(zw=3\sqrt2\,\mathrm{cis}\frac{3\pi}4=3\sqrt2\left(-\frac{\sqrt2}2+\frac{\sqrt2}2i\right)=-3+3i\).