The Exponential Form of a Complex Number

The exponential series, fed an imaginary number

The Maclaurin series for $e^x$eˣ is $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + …, and it adds up to $e^x$eˣ for every real number x. Its terms use only multiplying and adding, which work for complex numbers too, so the same series gives a meaning to e raised to an imaginary power.

Substitute $x=i\theta$x = iθ, where $\theta$θ is real: $e^{i\theta }=1+i\theta +\frac{{\left(i\theta \right)}^2}{2!}+\frac{{\left(i\theta \right)}^3}{3!}+\frac{{\left(i\theta \right)}^4}{4!}+\dots$e^(iθ) = 1 + iθ + (iθ)²/2! + (iθ)³/3! + (iθ)⁴/4! + …

The powers of i sort the terms

The powers of i repeat every four: $i^0=1$i⁰ = 1, $i^1=i$i¹ = i, $i^2=-1$i² = −1, $i^3=-i$i³ = −i, and then $i^4=1$i⁴ = 1 again. So ${\left(i\theta \right)}^2=-{\theta }^2$(iθ)² = −θ², ${\left(i\theta \right)}^3=-i{\theta }^3$(iθ)³ = −iθ³, ${\left(i\theta \right)}^4={\theta }^4$(iθ)⁴ = θ⁴ and ${\left(i\theta \right)}^5=i{\theta }^5$(iθ)⁵ = iθ⁵.

The series becomes $1+i\theta -\frac{{\theta }^2}{2!}-\frac{i{\theta }^3}{3!}+\frac{{\theta }^4}{4!}+\frac{i{\theta }^5}{5!}-\dots$1 + iθ − θ²/2! − iθ³/3! + θ⁴/4! + iθ⁵/5! − … The even powers have no i and the odd powers have one, so collect them separately: $e^{i\theta }=\left(1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\dots \right)+i\left(\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-\dots \right)$e^(iθ) = (1 − θ²/2! + θ⁴/4! − …) + i(θ − θ³/3! + θ⁵/5! − …).

Cosine and sine

The bracket without i is the Maclaurin series for $\cos \theta$cos θ, and the bracket with i is the series for $\sin \theta$sin θ. So $e^{i\theta }=\cos \theta +i\sin \theta$e^(iθ) = cos θ + i sin θ.

From here on the angle is in RADIANS. The series $\cos \theta =1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\dots$cos θ = 1 − θ²/2! + θ⁴/4! − … and $\sin \theta =\theta -\frac{{\theta }^3}{3!}+\dots$sin θ = θ − θ³/3! + … hold only for $\theta$θ in radians, so $e^{i\theta }=\cos \theta +i\sin \theta$e^(iθ) = cos θ + i sin θ does too. A half turn is $\pi$π radians, so 30° is $\frac{\pi }{6}$π/6, 60° is $\frac{\pi }{3}$π/3, 90° is $\frac{\pi }{2}$π/2 and 180° is $\pi$π.

Check at $\theta =1$θ = 1, which is 1 radian, about 57.3°. The first four terms of each bracket give $1-\frac{1}{2}+\frac{1}{24}-\frac{1}{720}\approx 0.5403$1 − 1/2 + 1/24 − 1/720 ≈ 0.5403 and $1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}\approx 0.8415$1 − 1/6 + 1/120 − 1/5040 ≈ 0.8415, and $\cos 1\approx 0.5403$cos 1 ≈ 0.5403, $\sin 1\approx 0.8415$sin 1 ≈ 0.8415.

The terms of the series for $e^{i\theta }$e^(iθ) at $\theta =1$θ = 1, laid head to tail: 1, then i, then $-\frac{1}{2}$−1/2, then $-\frac{i}{6}$−i/6, then $\frac{1}{24}$1/24. Each turns a quarter turn from the one before, and from the third on each is shorter; their total closes on the point $\cos 1+i\sin 1\approx 0.540+0.841i$cos 1 + i sin 1 ≈ 0.540 + 0.841i, on the dashed unit circle.

A point on the unit circle

The length of $e^{i\theta }$e^(iθ) is $\sqrt{{\cos }^2\theta +{\sin }^2\theta }=1$√(cos²θ + sin²θ) = 1, and its argument is $\theta$θ. So $e^{i\theta }$e^(iθ) is the point on the unit circle at angle $\theta$θ from the positive real axis.

The quarter turns: $e^{i\cdot 0}=1$e^(i·0) = 1, $e^{\frac{i\pi }{2}}=\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)=i$e^(iπ/2) = cos(π/2) + i sin(π/2) = i, $e^{i\pi }=\cos \pi +i\sin \pi =-1$e^(iπ) = cos π + i sin π = −1, and $e^{-\frac{i\pi }{2}}=-i$e^(−iπ/2) = −i. Written another way, $e^{i\pi }+1=0$e^(iπ) + 1 = 0. A full turn returns to the start: $e^{2\pi i}=1$e^(2πi) = 1.

$e^{\frac{i\pi }{3}}$e^(iπ/3) is the point at 60°: $\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)=\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i\approx 0.5+0.866i$cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i ≈ 0.5 + 0.866i.

Five values of $e^{i\theta }$e^(iθ) on the unit circle: 1 at $\theta =0$θ = 0, $\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i$1/2 + (√3/2)i at $\frac{\pi }{3}$π/3, i at $\frac{\pi }{2}$π/2, −1 at $\pi$π and −i at $-\frac{\pi }{2}$−π/2.

Every complex number as re${}^{i\theta }$^(iθ)

A complex number with modulus r and argument $\theta$θ is $z=r\left(\cos \theta +i\sin \theta \right)$z = r(cos θ + i sin θ), and the bracket is $e^{i\theta }$e^(iθ). So z = re${}^{i\theta }$^(iθ): r is the distance from the origin, and $\theta$θ, in radians, is the direction.

$2\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)=2e^{\frac{i\pi }{3}}$2(cos(π/3) + i sin(π/3)) = 2e^(iπ/3), which is $1+\sqrt{3}i$1 + √3 i. 1 + i has modulus $\sqrt{2}$√2 and argument $\frac{\pi }{4}$π/4, so $1+i=\sqrt{2}{e}^{\frac{i\pi }{4}}$1 + i = √2 e^(iπ/4). $1-i=\sqrt{2}{e}^{-\frac{i\pi }{4}}$1 − i = √2 e^(−iπ/4), its argument below the axis.

On the axes: $5=5e^{i\cdot 0}$5 = 5e^(i·0), $3i=3e^{\frac{i\pi }{2}}$3i = 3e^(iπ/2), $-3=3e^{i\pi }$−3 = 3e^(iπ) and $-2i=2e^{-\frac{i\pi }{2}}$−2i = 2e^(−iπ/2). Arguments are taken in $-\pi <\theta \le \pi$−π < θ ≤ π, the radian form of $-180^{\circ }<\theta \le 180^{\circ }$−180° < θ ≤ 180°.

Lengths multiply, angles add

Index laws add powers: $e^a\times e^b=e^{a+b}$e^a × e^b = e^(a + b). So $r_1{e}^{i{\theta }_1}\times r_2{e}^{i{\theta }_2}=r_1{r}_2{e}^{i\left({\theta }_1+{\theta }_2\right)}$r₁e^(iθ₁) × r₂e^(iθ₂) = r₁r₂ e^(i(θ₁ + θ₂)). The lengths multiply and the angles add, the rule for multiplying complex numbers met in polar form.

$2e^{\frac{i\pi }{6}}\times 3e^{\frac{i\pi }{3}}=6e^{i\left(\frac{\pi }{6}+\frac{\pi }{3}\right)}=6e^{\frac{i\pi }{2}}=6i$2e^(iπ/6) × 3e^(iπ/3) = 6e^(i(π/6 + π/3)) = 6e^(iπ/2) = 6i. In parts, $2e^{\frac{i\pi }{6}}=\sqrt{3}+i$2e^(iπ/6) = √3 + i and $3e^{\frac{i\pi }{3}}=\frac{3}{2}+\left(\frac{3\sqrt{3}}{2}\right)i$3e^(iπ/3) = 3/2 + (3√3/2)i, and $\left(\sqrt{3}+i\right)\left(\frac{3}{2}+\left(\frac{3\sqrt{3}}{2}\right)i\right)=\frac{3\sqrt{3}}{2}+\left(\frac{9}{2}\right)i+\left(\frac{3}{2}\right)i+\left(\frac{3\sqrt{3}}{2}\right)i^2=0+6i$(√3 + i)(3/2 + (3√3/2)i) = 3√3/2 + (9/2)i + (3/2)i + (3√3/2)i² = 0 + 6i.

Dividing subtracts the angles: $6e^{\frac{i\pi }{2}}/2e^{\frac{i\pi }{6}}=3e^{\frac{i\pi }{3}}$6e^(iπ/2) / 2e^(iπ/6) = 3e^(iπ/3). The conjugate of re${}^{i\theta }$^(iθ) is re${}^{-i\theta }$^(−iθ). And De Moivre’s theorem is the index law ${\left(e^{i\theta }\right)}^n=e^{(}$(e^(iθ))ⁿ = e^(in$\theta )$θ).

The usual mistakes

Moving r into the exponent. $2e^{\frac{i\pi }{3}}$2e^(iπ/3) is not $e^{\frac{2i\pi }{3}}$e^(2iπ/3): the second has length 1 and angle $\frac{2\pi }{3}$2π/3.

Dropping the i. $2e^{\frac{\pi }{3}}$2e^(π/3) is a real number, about 5.70; it is the i in the exponent that makes an angle.

Giving $e^{i\pi }=1$e^(iπ) = 1. $\pi$π is half a turn, so $e^{i\pi }=-1$e^(iπ) = −1; a full turn, $e^{2\pi i}$e^(2πi), is 1.

Adding the lengths, or multiplying the exponents. $2e^{\frac{i\pi }{6}}\times 3e^{\frac{i\pi }{3}}$2e^(iπ/6) × 3e^(iπ/3) is $6e^{\frac{i\pi }{2}}$6e^(iπ/2), not $5e^{\frac{i\pi }{2}}$5e^(iπ/2) or $6e^{\frac{i{\pi }^2}{18}}$6e^(iπ²/18).

Putting degrees in the exponent. $e^{60i}$e^(60i) is not $e^{\frac{i\pi }{3}}$e^(iπ/3): it is the point at 60 radians, about −0.952 − 0.305i.

Four loudspeakers slightly out of step

In the application below, the phasors 1, $e^{\frac{i\pi }{6}}$e^(iπ/6), $e^{\frac{2i\pi }{6}}$e^(2iπ/6) and $e^{\frac{3i\pi }{6}}$e^(3iπ/6) are 1 at 0°, 30°, 60° and 90°. Their sum is the sum drawn in Summing a Series with De Moivre at $\theta =30^{\circ }$θ = 30°: $\frac{3+\sqrt{3}}{2}+\left(\frac{3+\sqrt{3}}{2}\right)i\approx 2.366+2.366i$(3 + √3)/2 + ((3 + √3)/2)i ≈ 2.366 + 2.366i, which has length about 3.35 and argument $\frac{\pi }{4}$π/4, or 45°.