IB

View all Questions

Question

Mathematics

How do you raise a complex number in exponential form to powers?

Answer

Mathematics

Expert Answer

If z=reiθz=re^{i\theta}, raise the modulus, the distance from the origin, to the required power and multiply the angle by that power:

zn=rneinθ\boxed{z^n=r^ne^{in\theta}}

This is De Moivre's theorem in exponential form. For a positive integer nn, it follows from multiplying nn copies of reiθre^{i\theta}. Negative integer powers also work when z≠0z\ne0.

Example: square and cube a complex number

Let

z=2eiπ/3z=2e^{i\pi/3}

For the square, the modulus becomes 22=42^2=4 and the angle becomes 2π/32\pi/3:

z2=4e2πi/3z^2=4e^{2\pi i/3}

To express this as a+bia+bi, use eiϕ=cos⁡ϕ+isin⁡ϕe^{i\phi}=\cos\phi+i\sin\phi:

z2=4(cos⁡2π3+isin⁡2π3)=4(−12+32i)=−2+23 i\begin{aligned} z^2&=4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\\ &=4\left(-\frac12+\frac{\sqrt3}{2}i\right)\\ &=\boxed{-2+2\sqrt3\,i} \end{aligned}

For the cube,

z3=8eiπ=8(−1+0i)=−8z^3=8e^{i\pi}=8(-1+0i)=\boxed{-8}

The negative answer is consistent with an angle of π\pi: the point lies on the negative real axis.

Keep the modulus and angle separate

PowerModulusAngle before reductionCartesian form
zz22π/3\pi/31+3 i1+\sqrt3\,i
z2z^2442π/32\pi/3−2+23 i-2+2\sqrt3\,i
z3z^388π\pi−8-8
z4z^416164π/34\pi/3−8−83 i-8-8\sqrt3\,i

Unit-direction points for angles pi over 3, two pi over 3 and pi; these show direction only, not the increasing moduli of z's powers.

These points show w, w² and w³ on the unit circle: they show direction, not the growing distances of the powers of z.

The diagram uses w=z2=eiπ/3w=\dfrac{z}{2}=e^{i\pi/3} to show just the changes in direction. Its powers have modulus 1. To obtain znz^n, multiply the corresponding unit-direction point by 2n2^n.

Reduce the angle only when needed

Angles differing by 2π2\pi describe the same direction. For example, 4π/34\pi/3 and −2π/3-2\pi/3 are equivalent. If the question requests the principal argument, the angle in the specified interval (−π,π](-\pi,\pi], use −2π/3-2\pi/3.

Do not reduce the modulus when reducing the angle. Since 6(π/3)=2π6(\pi/3)=2\pi,

z6=64e2πi=64z^6=64e^{2\pi i}=64

not 1.

What about a negative power?

Take the reciprocal of the modulus and reverse the angle:

z−1=12e−iπ/3=14−34iz^{-1}=\frac12e^{-i\pi/3}=\frac14-\frac{\sqrt3}{4}i

Multiplying this by 1+3 i1+\sqrt3\,i gives 1, which checks the result. Use radian mode when evaluating angles written with π\pi.

Try the complex-number powers question in the AA HL Number and Algebra Questionbank. As you convert the square and cube to Cartesian form, keep track of the modulus and angle separately so that you account for both the change in distance and the rotation.

Answered by Revision Village IB Expert

Interested in diving deeper into this concept?

Explore More IB Mathematics Resources

Over 80% of IB students globally are experiencing the power of Revision Village