Representing complex numbers by length and angle — the gateway to elegant multiplication, division, and the most beautiful equation in mathematics
Every complex number $z = a + bi$ can be written in terms of its distance from the origin (modulus $r$) and angle from the positive real axis (argument $\theta$):
The angle $\theta$ is called the argument of $z$. The principal argument $\text{Arg}(z) \in (-\pi, \pi]$.
So $z = r\cos\theta + ir\sin\theta$.
Using $\theta = \arctan(b/a)$ alone only gives the correct angle when $a > 0$ (Quadrants I and IV). For $a < 0$ (Quadrants II and III), you must add $\pi$ to get the correct angle. Always sketch the point in the complex plane first.
Convert $z = -1 + \sqrt{3}\,i$ to polar form.
Modulus: $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$
Since $a = -1 < 0$ (Quadrant II): $\theta = \arctan\!\bigl(\frac{\sqrt{3}}{-1}\bigr) + \pi = \arctan(-\sqrt{3}) + \pi = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$
This connects the complex exponential to trigonometric functions and allows us to write the polar form compactly as $z = re^{i\theta}$.
Recall the Taylor series for $e^x$, $\cos x$, and $\sin x$:
Substituting $x = i\theta$ into the exponential series and using $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, etc., the even-powered terms give $\cos\theta$ and the odd-powered terms give $i\sin\theta$.
| $\theta$ | $e^{i\theta}$ | Value |
|---|---|---|
| $0$ | $e^{0}$ | $1$ |
| $\pi/2$ | $e^{i\pi/2}$ | $i$ |
| $\pi$ | $e^{i\pi}$ | $-1$ |
| $3\pi/2$ | $e^{3i\pi/2}$ | $-i$ |
| $2\pi$ | $e^{2\pi i}$ | $1$ |
Setting $\theta = \pi$: $e^{i\pi} + 1 = 0$. This single equation links the five most fundamental constants in mathematics: $e$, $i$, $\pi$, $1$, and $0$.
From Euler's formula and its conjugate $e^{-i\theta} = \cos\theta - i\sin\theta$:
When you solve $y'' + \omega^2 y = 0$, the characteristic roots are $r = \pm i\omega$, giving $y = c_1 e^{i\omega t} + c_2 e^{-i\omega t}$. Using the formulas above, this becomes $y = A\cos\omega t + B\sin\omega t$ with real constants — the familiar harmonic oscillator solution.
The polar form makes multiplication and division beautifully simple:
If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, then:
Multiply: multiply moduli, add arguments.
Divide: divide moduli, subtract arguments.
Let $z_1 = 2e^{i\pi/6}$ and $z_2 = 3e^{i\pi/4}$. Find $z_1 z_2$ and $z_1/z_2$.
Compute $(1+i)^8$ using polar form.
Convert to polar: $1 + i = \sqrt{2}\, e^{i\pi/4}$
Much simpler than multiplying $(1+i)$ eight times!
In differential equations, the characteristic roots are often of the form $r = \alpha + \beta i$. The solution $e^{rt}$ then becomes:
The factor $e^{\alpha t}$ is the amplitude envelope (growth if $\alpha > 0$, decay if $\alpha < 0$). The factor $\cos\beta t + i\sin\beta t$ is the oscillatory part with angular frequency $\beta$.
If a characteristic equation has roots $r = -2 \pm 3i$, the general solution is:
This is an oscillation with frequency $3$ rad/s that decays exponentially with time constant $1/2$ seconds. The amplitude halves roughly every $0.35$ seconds.
In AC circuit analysis, the steady-state response at frequency $\omega$ is $\text{Re}[V_0 e^{i(\omega t + \phi)}] = V_0 \cos(\omega t + \phi)$. The complex exponential carries both amplitude ($V_0$) and phase ($\phi$) — this is the phasor representation.
Step 1: Find the modulus
$$r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$Step 2: Locate the quadrant before touching $\arctan$
Here $a = -1 < 0$ and $b = \sqrt{3} > 0$, so $z$ lies in Quadrant II. Its argument must land between $\pi/2$ and $\pi$.
Step 3: Compute the reference angle
$$\arctan\!\left|\frac{b}{a}\right| = \arctan(\sqrt{3}) = \frac{\pi}{3}$$A calculator asked for $\arctan(\sqrt{3}/(-1)) = \arctan(-\sqrt{3})$ would report $-\pi/3$, which sits in Quadrant IV — the wrong place.
Step 4: Adjust for Quadrant II
$$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$Step 5: Write both polar forms
$$\boxed{z = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right) = 2e^{i2\pi/3}}$$Step 1: Multiply moduli, add arguments
$$z_1 z_2 = (2)(3)\,e^{i(\pi/3 + \pi/6)} = 6e^{i\pi/2}$$Step 2: Convert the product back to rectangular form
$$6e^{i\pi/2} = 6\left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right) = 6(0 + i) = 6i$$Step 3: Divide moduli, subtract arguments
$$\frac{z_1}{z_2} = \frac{2}{3}e^{i(\pi/3 - \pi/6)} = \frac{2}{3}e^{i\pi/6}$$Step 4: Convert the quotient back
$$\frac{2}{3}\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right) = \frac{2}{3}\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \frac{\sqrt{3}}{3} + \frac{1}{3}i$$ $$\boxed{z_1 z_2 = 6i, \qquad \frac{z_1}{z_2} = \frac{\sqrt{3}}{3} + \frac{1}{3}i}$$Step 1: Put the base in polar form
$1 + i$ has modulus $\sqrt{1 + 1} = \sqrt{2}$ and, being in Quadrant I with equal parts, argument $\pi/4$:
$$1 + i = \sqrt{2}\,e^{i\pi/4}$$Step 2: Raise to the eighth power — modulus and argument separate
$$(1+i)^8 = \left(\sqrt{2}\right)^8 e^{i(8 \cdot \pi/4)} = 2^{4}\,e^{i2\pi}$$Step 3: Reduce the angle
$e^{i2\pi} = \cos 2\pi + i\sin 2\pi = 1$, so
$$(1+i)^8 = 16 \cdot 1 = 16$$ $$\boxed{(1+i)^8 = 16}$$Step 1: Split the exponent
The exponential law $e^{u+v} = e^u e^v$ still holds for complex exponents:
$$e^{(-2+3i)t} = e^{-2t} \cdot e^{i3t}$$Step 2: Apply Euler's formula to the purely imaginary factor
$$e^{i3t} = \cos 3t + i\sin 3t$$Step 3: Combine
$$e^{(-2+3i)t} = e^{-2t}\cos 3t + i\,e^{-2t}\sin 3t$$Step 4: Read off the two real functions
$$\operatorname{Re} = e^{-2t}\cos 3t, \qquad \operatorname{Im} = e^{-2t}\sin 3t$$Note also that the modulus is $\left|e^{(-2+3i)t}\right| = e^{-2t}$, since $|e^{i3t}| = 1$ for every real $t$. The real part $-2$ controls the envelope; the imaginary part $3$ controls only the frequency.
$$\boxed{e^{(-2+3i)t} = e^{-2t}\left(\cos 3t + i \sin 3t\right)}$$These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.
Step 1: Modulus
$$r = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$$Step 2: Argument, using the quadrant
Both parts are negative, so $z$ is in Quadrant III. The reference angle is $\arctan(2/2) = \pi/4$, and the principal argument in Quadrant III is
$$\theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$$ $$z = 2\sqrt{2}\,e^{-i3\pi/4}$$Step 3: Apply De Moivre
$$z^5 = \left(2\sqrt{2}\right)^5 e^{-i15\pi/4}$$The modulus is $\left(2^{3/2}\right)^5 = 2^{15/2} = 2^7\sqrt{2} = 128\sqrt{2}$.
Step 4: Reduce the angle modulo $2\pi$
$$-\frac{15\pi}{4} + 4\pi = \frac{\pi}{4} \;\Longrightarrow\; z^5 = 128\sqrt{2}\,e^{i\pi/4}$$Step 5: Back to rectangular form
$$128\sqrt{2}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) = 128 + 128i$$ $$\boxed{z = 2\sqrt{2}\,e^{-i3\pi/4}, \qquad z^5 = 128\sqrt{2}\,e^{i\pi/4} = 128 + 128i}$$Step 1: Write the same number two ways
$$e^{i3\theta} = \cos 3\theta + i\sin 3\theta \qquad\text{and}\qquad e^{i3\theta} = \left(e^{i\theta}\right)^3 = (\cos\theta + i\sin\theta)^3$$Step 2: Expand the cube
$$(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta + 3i^2\cos\theta\sin^2\theta + i^3\sin^3\theta$$Using $i^2 = -1$ and $i^3 = -i$ and grouping:
$$= \left(\cos^3\theta - 3\cos\theta\sin^2\theta\right) + i\left(3\cos^2\theta\sin\theta - \sin^3\theta\right)$$Step 3: Match real parts
$$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$$Step 4: Eliminate $\sin^2\theta$ using $\sin^2\theta = 1 - \cos^2\theta$
$$\cos 3\theta = \cos^3\theta - 3\cos\theta\left(1 - \cos^2\theta\right) = \cos^3\theta - 3\cos\theta + 3\cos^3\theta$$ $$\boxed{\cos 3\theta = 4\cos^3\theta - 3\cos\theta}$$Step 1: Solve the characteristic equation
$$r^2 - 2r + 17 = 0 \;\Longrightarrow\; r = \frac{2 \pm \sqrt{4 - 68}}{2} = \frac{2 \pm \sqrt{-64}}{2} = \frac{2 \pm 8i}{2} = 1 \pm 4i$$Step 2: Write the complex solution and split it with Euler
$$e^{(1+4i)t} = e^{t}e^{i4t} = e^{t}\left(\cos 4t + i \sin 4t\right)$$Step 3: Take real and imaginary parts
Because the equation has real coefficients, the real and imaginary parts of any complex solution are each real solutions:
$$y_1 = e^{t}\cos 4t, \qquad y_2 = e^{t}\sin 4t$$Step 4: Combine into the general real solution
$$\boxed{y(t) = e^{t}\left(c_1\cos 4t + c_2 \sin 4t\right)}$$Step 5: Interpret
$\alpha = 1 > 0$ means the envelope $e^{t}$ grows, so this is an unstable oscillation; $\beta = 4$ means the oscillation has angular frequency $4$ rad/s, i.e. period $2\pi/4 = \pi/2$.
Convert $z = -\sqrt{3} - i$ to polar form $re^{i\theta}$ with $\theta \in (-\pi, \pi]$.
$r = \sqrt{3 + 1} = 2$. The point is in Quadrant III ($a < 0, b < 0$).
$\arctan\!\bigl(\frac{-1}{-\sqrt{3}}\bigr) = \arctan\!\bigl(\frac{1}{\sqrt{3}}\bigr) = \frac{\pi}{6}$, but we need to add $\pi$ and then subtract $2\pi$ to get principal argument:
$\theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$. For principal argument: $\frac{7\pi}{6} - 2\pi = -\frac{5\pi}{6}$.
Use Euler's formula to prove that $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$.
Since $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$:
Write $e^{(1+2i)t}$ in the form $e^{\alpha t}(A\cos\beta t + B\sin\beta t)$ and identify $\alpha$ and $\beta$.
So $\alpha = 1$ (exponential growth) and $\beta = 2$ (oscillation frequency 2 rad/s).
Test your understanding with these multiple-choice questions. Click a choice to see if you're correct.
Answer: (B) $2e^{i2\pi/3}$. The modulus is $\sqrt{1+3} = 2$, and $z$ lies in Quadrant II, so $\theta = \pi - \pi/3 = 2\pi/3$. Option (A) is $1 + i\sqrt{3}$ (Quadrant I), option (C) is $-1 - i\sqrt{3}$ (Quadrant III), and option (D) has the wrong modulus — $r^2 = 4$, not $r = 4$.
Answer: (D) $8e^{i5\pi/12}$. Multiply the moduli ($4 \cdot 2 = 8$) and add the arguments ($\pi/6 + \pi/4 = 2\pi/12 + 3\pi/12 = 5\pi/12$). Option (A) multiplies the angles, option (B) adds the moduli, and option (C) is the quotient $z_1/z_2$.
Answer: (A) $-i$. $e^{i3\pi/2} = \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = 0 + i(-1) = -i$. Three quarter-turns counter-clockwise from $1$ lands on the negative imaginary axis — the same place as one quarter-turn clockwise, $e^{-i\pi/2}$.
Answer: (C) $16$. Since $1 + i = \sqrt{2}e^{i\pi/4}$, we get $(1+i)^8 = (\sqrt{2})^8 e^{i2\pi} = 16 \cdot 1 = 16$. Option (D) forgets that the modulus is $\sqrt{2}$, not $2$; options (A) and (B) misreduce the angle $8 \cdot \pi/4 = 2\pi$, which is a full turn back to the positive real axis.
Answer: (B) $e^{-3t}$. Split the exponent: $e^{(-3+2i)t} = e^{-3t}e^{i2t}$, and $\left|e^{i2t}\right| = \sqrt{\cos^2 2t + \sin^2 2t} = 1$. Only the real part of the exponent affects the size. Option (A) would be right only if the exponent were purely imaginary.
Answer: (D) $e^{2t}(\cos 5t - i\sin 5t)$. $e^{(2-5i)t} = e^{2t}e^{-i5t}$, and $e^{-i5t} = \cos(-5t) + i\sin(-5t) = \cos 5t - i \sin 5t$ because cosine is even and sine is odd. Option (B) swaps the roles of the real and imaginary parts, and (C) keeps only the real part.
You now have all the tools — rectangular and polar forms, Euler's formula, and the complex exponential. The next section puts it all together: finding $n$th roots via De Moivre's theorem and, most importantly, converting complex characteristic roots into real-valued general solutions of differential equations.
Drag the point, or use the sliders, to see how rectangular and polar form describe the same number. Watch what happens to the angle when you cross into a new quadrant — this is the single most common source of errors when converting to polar form.