CP 2.9 - Exponential form of a complex number

CP 2.9 - Exponential form of a complex number

An exponential can describe a turn as neatly as it describes growth. You will use the definition of eiθ\mathrm{e}^{\mathrm{i}\theta} to write complex numbers in exponential form, calculate with them, and recover sine and cosine from pairs of exponentials.

An exponential on the unit circle

Can an exponential equal −1-1, even though ex\mathrm{e}^x is positive for every real xx?

The crucial word is “real”. An imaginary exponent has a different job. Start at 11 on the unit circle and turn anticlockwise through θ\theta radians. The point you reach has coordinates (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta), so its complex number is cos⁡θ+isin⁡θ\cos\theta+\mathrm{i}\sin\theta. We give that number the shorter name eiθ\mathrm{e}^{\mathrm{i}\theta}.

You already know

A non-zero complex number can be written as r(cos⁡θ+isin⁡θ)r(\cos\theta+\mathrm{i}\sin\theta), where r=∣z∣>0r=|z|>0. Multiplication multiplies moduli and adds arguments; division divides moduli and subtracts arguments.

A quarter-turn gives eiπ/2=i\mathrm{e}^{\mathrm{i}\pi/2}=\mathrm{i}; a half-turn gives eiπ=−1\mathrm{e}^{\mathrm{i}\pi}=-1. The exponential has moved round a circle, rather than along the positive real axis.

[DIAGRAM: asset_name: Imaginary exponentials on the unit circle; asset_slug: 9fm0-cp12-unit-exponentials; caption: The exponents give angles and the points stay one unit from the origin; description: A unit circle on equal-scale Argand axes has the points 1, i, minus 1 and minus i marked. The point exp(i pi over 3) at one half plus i root 3 over 2 and its conjugate exp(minus i pi over 3) are highlighted. A dashed vertical projection meets the real axis at one half. Anticlockwise and clockwise angle arcs are labelled pi over 3 and minus pi over 3. A side panel states modulus equals 1 and adding 2 pi to an angle returns to the same point]
The exponents give angles and the points stay one unit from the origin

The definition uses radians. It agrees with the multiplication rule you already know, which is why exponential notation is so useful.

Imaginary exponential

For real θ\theta,

eiθ=cos⁡θ+isin⁡θ\mathrm{e}^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta

For example, eiπ/3=12+i32\mathrm{e}^{\mathrm{i}\pi/3}=\frac12+\frac{\mathrm{i}\sqrt3}{2}. It does not mean e×i×θ\mathrm{e}\times\mathrm{i}\times\theta.

The modulus is cos⁡2θ+sin⁡2θ=1\sqrt{\cos^2\theta+\sin^2\theta}=1. Adding any whole number of full turns changes the angle label but leaves the number unchanged: ei(θ+2kπ)=eiθ\mathrm{e}^{\mathrm{i}(\theta+2k\pi)}=\mathrm{e}^{\mathrm{i}\theta} for integer kk. In particular, e2027πi=−1\mathrm{e}^{2027\pi\mathrm{i}}=-1, even though the exponent looks enormous: 2027π=π+1013(2π)2027\pi=\pi+1013(2\pi).

Key idea

In eiθ\mathrm{e}^{\mathrm{i}\theta}, the real number θ\theta specifies a turn; the modulus is always 11.

Read the exponent as an angle before deciding where the point lies.

Converting between Cartesian and exponential form

The exponential form packages the same modulus and argument you have already been finding into a shorter expression.

For example, 1+i31+\mathrm{i}\sqrt3 has modulus 22 and argument π/3\pi/3, so it is 2eiπ/32\mathrm{e}^{\mathrm{i}\pi/3}. The coefficient supplies the distance from the origin; the exponent supplies the direction.

Exponential form

For z=x+iy≠0z=x+\mathrm{i}y\ne0, choose an argument θ\theta and let r=x2+y2>0r=\sqrt{x^2+y^2}>0. Then

z=reiθ=r(cos⁡θ+isin⁡θ)z=r\mathrm{e}^{\mathrm{i}\theta} =r(\cos\theta+\mathrm{i}\sin\theta)

The principal argument satisfies −π<θ≤π-\pi<\theta\le\pi, unless a question requests another interval. Not in the formula booklet: learn it.

Zero has no argument and therefore no exponential form with r>0r>0. For any other number, locate the quadrant before using an inverse tangent.

Worked example 1: a point above the real axis

Question. Express −3+i-\sqrt3+\mathrm{i} in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi.

Plan. Find the modulus, then use the second quadrant to select the argument.

r=(−3)2+12=2distance from 0tan⁡α=13,α=π6reference angleθ=π−π6=5π6second quadrant\begin{aligned} r&=\sqrt{(-\sqrt3)^2+1^2}=2 && \text{distance from }0 \\ \tan\alpha&=\frac1{\sqrt3},\quad \alpha=\frac\pi6 && \text{reference angle} \\ \theta&=\pi-\frac\pi6=\frac{5\pi}6 && \text{second quadrant} \end{aligned}

Why this step? The inverse tangent of −1/3-1/\sqrt3 alone gives a fourth-quadrant angle, whereas the negative real part puts this point on the left.

Answer. z=2e5πi/6z=2\mathrm{e}^{5\pi\mathrm{i}/6}.

Check. 2(cos⁡(5π/6)+isin⁡(5π/6))=2(−3/2+i/2)=−3+i2(\cos(5\pi/6)+\mathrm{i}\sin(5\pi/6))=2(-\sqrt3/2+\mathrm{i}/2)=-\sqrt3+\mathrm{i}.

Worked example 2 changes only the sign of the imaginary part. The modulus stays fixed, but the principal argument changes sign.

Worked example 2: reflecting below the real axis

Question. Express −3−i-\sqrt3-\mathrm{i} in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi.

Plan. Keep the modulus and reference angle, then measure clockwise to the third-quadrant point.

r=(−3)2+(−1)2=2same squared coordinatesα=π6same reference angleθ=−π+π6=−5π6principal third-quadrant angle\begin{aligned} r&=\sqrt{(-\sqrt3)^2+(-1)^2}=2 && \text{same squared coordinates} \\ \alpha&=\frac\pi6 && \text{same reference angle} \\ \theta&=-\pi+\frac\pi6=-\frac{5\pi}6 && \text{principal third-quadrant angle} \end{aligned}

Why this step? 7π/67\pi/6 reaches the right point, but it is outside the requested interval. Subtracting 2π2\pi gives −5π/6-5\pi/6.

Answer. z=2e−5πi/6z=2\mathrm{e}^{-5\pi\mathrm{i}/6}.

Check. 2cos⁡(−5π/6)=−32\cos(-5\pi/6)=-\sqrt3 and 2sin⁡(−5π/6)=−12\sin(-5\pi/6)=-1, giving the required Cartesian parts.

The positive modulus makes the two forms easy to compare: a sign change in a coordinate belongs in the angle, not in a negative value of rr.

Watch out

Wrong: −3−i=2eiπ/6-\sqrt3-\mathrm{i}=2\mathrm{e}^{\mathrm{i}\pi/6}. Matching the tangent is insufficient: π/6\pi/6 gives two positive coordinates. Right: locate the point first, then write 2e−5πi/62\mathrm{e}^{-5\pi\mathrm{i}/6}.

Change the distance while keeping the direction in your turn.

Products, quotients and powers in exponential form

Exponential notation makes the modulus-argument rules look like the index laws they describe.

For instance, eiπ/3eiπ/6=eiπ/2=i\mathrm{e}^{\mathrm{i}\pi/3}\mathrm{e}^{\mathrm{i}\pi/6}=\mathrm{e}^{\mathrm{i}\pi/2}=\mathrm{i}: two successive turns combine into a quarter-turn. This follows from the established multiplication rule, so it does not require a new rule for complex multiplication.

Calculating with exponential form

For positive real r,sr,s, real θ,ϕ\theta,\phi, and integer nn:

(reiθ)(seiϕ)=rsei(θ+ϕ)reiθseiϕ=rsei(θ−ϕ)(reiθ)n=rneinθ\begin{aligned} (r\mathrm{e}^{\mathrm{i}\theta})(s\mathrm{e}^{\mathrm{i}\phi})&=rs\mathrm{e}^{\mathrm{i}(\theta+\phi)} \\ \frac{r\mathrm{e}^{\mathrm{i}\theta}}{s\mathrm{e}^{\mathrm{i}\phi}}&=\frac rs\mathrm{e}^{\mathrm{i}(\theta-\phi)} \\ (r\mathrm{e}^{\mathrm{i}\theta})^n&=r^n\mathrm{e}^{\mathrm{i}n\theta} \end{aligned}

The power rule is de Moivre's theorem written exponentially; its trigonometric form is in the formula booklet. The exponential calculation rules above are not printed there. Not in the formula booklet: learn it.

For a principal argument, reduce the final angle after calculating. Reducing it earlier is also valid, but usually creates extra work.

Worked example 3: a product with a large argument

Question. Given z=2e5πi/6z=2\mathrm{e}^{5\pi\mathrm{i}/6} and w=3eπi/2w=3\mathrm{e}^{\pi\mathrm{i}/2}, express zwzw in Cartesian form.

Plan. Multiply moduli, add arguments, then use exact trigonometric values.

zw=6ei(5π/6+π/2)multiply and add=6e4πi/3=6e−2πi/3subtract 2π=6(−12−i32)convert to Cartesian form=−3−3i3\begin{aligned} zw&=6\mathrm{e}^{\mathrm{i}(5\pi/6+\pi/2)} && \text{multiply and add} \\ &=6\mathrm{e}^{4\pi\mathrm{i}/3}=6\mathrm{e}^{-2\pi\mathrm{i}/3} && \text{subtract }2\pi \\ &=6\left(-\frac12-\frac{\mathrm{i}\sqrt3}{2}\right) && \text{convert to Cartesian form} \\ &=-3-3\mathrm{i}\sqrt3 \end{aligned}

Why this step? Reducing 4π/34\pi/3 to −2π/3-2\pi/3 identifies the principal argument without changing the number.

Answer. zw=−3−3i3zw=-3-3\mathrm{i}\sqrt3.

Check. Directly, z=−3+iz=-\sqrt3+\mathrm{i} and w=3iw=3\mathrm{i}, so zw=−3−3i3zw=-3-3\mathrm{i}\sqrt3; its modulus is 66.

Worked example 4 changes the operation from multiplication to division, using the same two numbers. Subtract the denominator's argument from the numerator's.

Worked example 4: a quotient in exact form

Question. For the same zz and ww, find z/wz/w in Cartesian form, showing your working.

Plan. Divide the moduli and subtract the arguments in the order of the quotient.

zw=23ei(5π/6−π/2)divide and subtract=23eπi/3combine the angles=23(12+i32)use the definition=13+i33\begin{aligned} \frac zw&=\frac23\mathrm{e}^{\mathrm{i}(5\pi/6-\pi/2)} && \text{divide and subtract} \\ &=\frac23\mathrm{e}^{\pi\mathrm{i}/3} && \text{combine the angles} \\ &=\frac23\left(\frac12+\frac{\mathrm{i}\sqrt3}{2}\right) && \text{use the definition} \\ &=\frac13+\frac{\mathrm{i}\sqrt3}{3} \end{aligned}

Why this step? Division reverses the turn contributed by ww; adding the angles would calculate a product instead.

Answer. z/w=13+i33z/w=\frac13+\frac{\mathrm{i}\sqrt3}{3}.

Check. Multiplying by 3i3\mathrm{i} gives i−3=z\mathrm{i}-\sqrt3=z, as required.

A power repeats a product: (2eπi/3)3=8eπi=−8(2\mathrm{e}^{\pi\mathrm{i}/3})^3=8\mathrm{e}^{\pi\mathrm{i}}=-8. The coefficient is cubed while the angle is tripled.

Watch out

Wrong: (2eiθ)3=6e3iθ(2\mathrm{e}^{\mathrm{i}\theta})^3=6\mathrm{e}^{3\mathrm{i}\theta}. The modulus is raised to the power too. Right: (2eiθ)3=8e3iθ(2\mathrm{e}^{\mathrm{i}\theta})^3=8\mathrm{e}^{3\mathrm{i}\theta}.

In an exact-form question, a decimal Cartesian approximation is not a substitute for the requested surds. Show the exponential calculation before evaluating the trigonometric functions.

Recovering cosine and sine from exponentials

Pairing opposite turns separates the real and imaginary parts of an exponential.

Replacing θ\theta by −θ-\theta in the definition gives

e−iθ=cos⁡θ−isin⁡θ\mathrm{e}^{-\mathrm{i}\theta}=\cos\theta-\mathrm{i}\sin\theta

because cosine is even and sine is odd. Add the two definitions and the imaginary terms cancel; subtract them and the real terms cancel.

Cosine and sine as exponentials

For real θ\theta:

cos⁡θ=eiθ+e−iθ2,sin⁡θ=eiθ−e−iθ2i\cos\theta=\frac{\mathrm{e}^{\mathrm{i}\theta}+\mathrm{e}^{-\mathrm{i}\theta}}2, \qquad \sin\theta=\frac{\mathrm{e}^{\mathrm{i}\theta}-\mathrm{e}^{-\mathrm{i}\theta}}{2\mathrm{i}}

Also e−iθ=1/eiθ=(eiθ)∗\mathrm{e}^{-\mathrm{i}\theta}=1/\mathrm{e}^{\mathrm{i}\theta}=(\mathrm{e}^{\mathrm{i}\theta})^*. The reciprocal equals the conjugate here because the modulus is 11. Not in the formula booklet: learn it.

The same identities work for any real angle in place of θ\theta. If q=eiθq=\mathrm{e}^{\mathrm{i}\theta}, then qm+q−m=2cos⁡mθq^m+q^{-m}=2\cos m\theta for integer mm. This is the exponential version of the reciprocal pairing used in multiple-angle work.

Worked example 5: a sum of opposite turns

Question. Given q=eiθq=\mathrm{e}^{\mathrm{i}\theta}, show that q3+q−3=2cos⁡3θq^3+q^{-3}=2\cos3\theta.

Plan. Write both powers as exponentials and apply the definition at angles 3θ3\theta and −3θ-3\theta.

q3+q−3=e3iθ+e−3iθuse the power rule=(cos⁡3θ+isin⁡3θ)+(cos⁡3θ−isin⁡3θ)opposite angles=2cos⁡3θimaginary parts cancel\begin{aligned} q^3+q^{-3}&=\mathrm{e}^{3\mathrm{i}\theta}+\mathrm{e}^{-3\mathrm{i}\theta} && \text{use the power rule} \\ &=(\cos3\theta+\mathrm{i}\sin3\theta)+(\cos3\theta-\mathrm{i}\sin3\theta) && \text{opposite angles} \\ &=2\cos3\theta && \text{imaginary parts cancel} \end{aligned}

Why this step? The negative power is a reciprocal, hence an opposite angle. It is not a minus sign in front of q3q^3.

Answer. q3+q−3=2cos⁡3θq^3+q^{-3}=2\cos3\theta, as required.

Check. At θ=π/6\theta=\pi/6, q3=iq^3=\mathrm{i} and q−3=−iq^{-3}=-\mathrm{i}; both sides equal 00.

Worked example 6 changes only the plus sign to a minus sign. Now the real parts cancel instead.

Worked example 6: a difference of opposite turns

Question. Given q=eiθq=\mathrm{e}^{\mathrm{i}\theta}, show that q3−q−3=2isin⁡3θq^3-q^{-3}=2\mathrm{i}\sin3\theta.

Plan. Expand the same two exponentials, taking care to subtract the whole second bracket.

q3−q−3=e3iθ−e−3iθuse the power rule=(cos⁡3θ+isin⁡3θ)−(cos⁡3θ−isin⁡3θ)subtract both terms=2isin⁡3θreal parts cancel\begin{aligned} q^3-q^{-3}&=\mathrm{e}^{3\mathrm{i}\theta}-\mathrm{e}^{-3\mathrm{i}\theta} && \text{use the power rule} \\ &=(\cos3\theta+\mathrm{i}\sin3\theta)-(\cos3\theta-\mathrm{i}\sin3\theta) && \text{subtract both terms} \\ &=2\mathrm{i}\sin3\theta && \text{real parts cancel} \end{aligned}

Why this step? Subtracting the negative imaginary part produces another positive isin⁡3θ\mathrm{i}\sin3\theta; the imaginary unit remains in the answer.

Answer. q3−q−3=2isin⁡3θq^3-q^{-3}=2\mathrm{i}\sin3\theta, as required.

Check. At θ=π/6\theta=\pi/6, the left side is i−(−i)=2i\mathrm{i}-(-\mathrm{i})=2\mathrm{i}, matching 2isin⁡(π/2)2\mathrm{i}\sin(\pi/2).

These identities extract real-valued sine and cosine from complex expressions. They also settle the opening question: θ=π\theta=\pi gives eiπ=−1\mathrm{e}^{\mathrm{i}\pi}=-1, or eiπ+1=0\mathrm{e}^{\mathrm{i}\pi}+1=0.

Watch out

Wrong: sin⁡θ=(eiθ−e−iθ)/2\sin\theta=(\mathrm{e}^{\mathrm{i}\theta}-\mathrm{e}^{-\mathrm{i}\theta})/2. The numerator is 2isin⁡θ2\mathrm{i}\sin\theta, so divide by 2i2\mathrm{i}. Without that imaginary unit, the expression is generally imaginary rather than real.

The next check changes the power from 33 to 44. Use the same opposite-angle cancellation and show the missing factor explicitly.

Quick Recap

  • eiθ=cos⁡θ+isin⁡θ\mathrm{e}^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta describes a point on the unit circle.
  • For z≠0z\ne0, z=reiθz=r\mathrm{e}^{\mathrm{i}\theta} records its positive modulus and an argument; angles differing by 2π2\pi give the same number.
  • Multiply moduli and add arguments; divide moduli and subtract arguments; raise the modulus and multiply the argument for a power.
  • Adding opposite exponentials gives 2cos⁡θ2\cos\theta; subtracting them gives 2isin⁡θ2\mathrm{i}\sin\theta.

An exponential can equal −1-1: the imaginary exponent iπ\mathrm{i}\pi describes the half-turn from 11 to −1-1.