Unit 7, lesson 87

Polar Form of Complex Numbers

How to write a nonzero complex number by its distance from zero and an angle, why that angle is not unique, which angle is the principal one, and how to convert back to rectangular form.

  • Theory part 173
  • Practice part 174
  • Complex Numbers in Polar Form

A complex number a+bia + bi sits at the point (a,b)(a, b) of the complex plane. Like any point, it can be located by how far it is from the origin and by the direction you face to see it. Written that way, complex numbers become much easier to multiply, divide and raise to powers than in rectangular form.

Key terms

Modulus

Definition The modulus of the complex number z=a+biz = a + bi, with aa and bb real, is the real number

∣z∣=a2+b2.\lvert z \rvert = \sqrt{a^2 + b^2}.

It is the distance from 00 to zz in the complex plane. Always ∣z∣≥0\lvert z \rvert \ge 0, and ∣z∣=0\lvert z \rvert = 0 only for z=0z = 0.

In plain words The modulus is the length of the arrow from the origin to zz. For a real number it is the usual absolute value.

Example For z=−5+12iz = -5 + 12i,

∣z∣=25+144=169=13.\lvert z \rvert = \sqrt{25 + 144} = \sqrt{169} = 13.

Argument

Definition Let z=a+biz = a + bi be a nonzero complex number. An argument of zz is any real number θ\theta with

a=∣z∣cos⁡θ,b=∣z∣sin⁡θ.a = \lvert z \rvert\cos\theta, \qquad b = \lvert z \rvert\sin\theta.

Geometrically, θ\theta is a directed angle in radians from the positive real axis to the ray from 00 through zz, counterclockwise positive. The set of all arguments of zz is written arg⁡z\arg z.

In plain words The argument is the direction of the arrow to zz, measured from the positive real axis the same way as an angle in polar coordinates.

Example The number z=−2+2iz = -2 + 2i has ∣z∣=8=22\lvert z \rvert = \sqrt{8} = 2\sqrt{2} and lies in the second quadrant on the line y=−xy = -x, so θ=3π4\theta = \frac{3\pi}{4} is an argument. Indeed 22cos⁡3π4=−22\sqrt{2}\cos\frac{3\pi}{4} = -2 and 22sin⁡3π4=22\sqrt{2}\sin\frac{3\pi}{4} = 2.

Polar representation

Definition Every nonzero complex number zz can be written as

z=r(cos⁡θ+isin⁡θ),z = r(\cos\theta + i\sin\theta),

where r=∣z∣>0r = \lvert z \rvert > 0 and θ\theta is an argument of zz. This is the polar form, or trigonometric form, of zz. Conversely, if z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta) with r>0r > 0 and θ\theta real, then r=∣z∣r = \lvert z \rvert and θ\theta is an argument of zz. Unlike polar coordinates of points, the polar form of a complex number requires a positive rr.

In plain words A length times a direction. The factor cos⁡θ+isin⁡θ\cos\theta + i\sin\theta has length 11 and points the right way, and rr stretches it to the right length.

Example For z=1−3 iz = 1 - \sqrt{3}\,i the modulus is 1+3=2\sqrt{1 + 3} = 2, and cos⁡θ=12\cos\theta = \frac{1}{2}, sin⁡θ=−32\sin\theta = -\frac{\sqrt{3}}{2} hold for θ=−π3\theta = -\frac{\pi}{3}, so

1−3 i=2(cos⁡(−π3)+isin⁡(−π3)).\begin{aligned} &1 - \sqrt{3}\,i \\[0.4em] &\quad = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right). \end{aligned}

An expression such as −2(cos⁡π6+isin⁡π6)-2(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) is not in polar form, because its first factor is negative. Turning the angle by π\pi fixes it, since

−2(cos⁡π6+isin⁡π6)=2(cos⁡7π6+isin⁡7π6).\begin{aligned} &{-2}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) \\[0.8em] &\quad = 2\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right). \end{aligned}

Cis notation

Definition For a real number θ\theta, the abbreviation cis⁡θ\operatorname{cis}\theta means

cis⁡θ=cos⁡θ+isin⁡θ.\operatorname{cis}\theta = \cos\theta + i\sin\theta.

So the polar form of a nonzero zz is z=rcis⁡θz = r\operatorname{cis}\theta with r=∣z∣r = \lvert z \rvert. Every number cis⁡θ\operatorname{cis}\theta has modulus 11, since cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1.

In plain words The letters stand for cosine, ii, sine. It is shorthand that saves writing the same bracket again and again.

Example The number 4cis⁡5π64\operatorname{cis}\frac{5\pi}{6} in rectangular form is

4cos⁡5π6+4isin⁡5π6=−23+2i.4\cos\frac{5\pi}{6} + 4i\sin\frac{5\pi}{6} = -2\sqrt{3} + 2i.

Nonunique arguments

Definition If θ\theta is one argument of a nonzero complex number zz, then the arguments of zz are exactly the numbers

θ+2kπ,k∈Z.\theta + 2k\pi, \qquad k \in \mathbb{Z}.

The reason is that cos⁡α=cos⁡β\cos\alpha = \cos\beta and sin⁡α=sin⁡β\sin\alpha = \sin\beta hold together exactly when α−β\alpha - \beta is an integer multiple of 2π2\pi. So for r>0r > 0, the numbers rcis⁡αr\operatorname{cis}\alpha and rcis⁡βr\operatorname{cis}\beta are equal exactly when α\alpha and β\beta are coterminal.

In plain words One more full turn, or one fewer, points the arrow the same way. Every nonzero complex number has infinitely many arguments, all 2π2\pi apart.

Example The number ii has the arguments π2\frac{\pi}{2}, 5π2\frac{5\pi}{2} and −3π2-\frac{3\pi}{2}, among infinitely many others, so

i=cis⁡π2=cis⁡5π2=cis⁡(−3π2).i = \operatorname{cis}\frac{\pi}{2} = \operatorname{cis}\frac{5\pi}{2} = \operatorname{cis}\left(-\frac{3\pi}{2}\right).

Principal argument

Definition The principal argument of a nonzero complex number zz, written Arg⁡z\operatorname{Arg} z with a capital A, is the unique argument of zz in the interval (−π,π](-\pi, \pi]. For z=a+bi≠0z = a + bi \ne 0 it equals atan2⁡(b,a)\operatorname{atan2}(b, a). Some books use the interval [0,2π)[0, 2\pi) instead, so check which convention a problem uses. In this course Arg⁡z\operatorname{Arg} z always lies in (−π,π](-\pi, \pi].

In plain words Out of all the arguments, take the one between −π-\pi and π\pi. Numbers above the real axis get positive principal arguments and numbers below it get negative ones. The positive real axis gets 00 and the negative real axis gets π\pi.

Example Here are four principal arguments.

zzPositionArg⁡z\operatorname{Arg} z
3+i\sqrt{3} + ifirst quadrantπ6\frac{\pi}{6}
−1−i-1 - ithird quadrant−3π4-\frac{3\pi}{4}
−2i-2inegative imaginary axis−π2-\frac{\pi}{2}
−1-1negative real axisπ\pi

With the other convention, the principal argument of −1−i-1 - i would be 5π4\frac{5\pi}{4} instead.

Zero has no unique argument

Definition The number 00 has modulus 00, and

0=0cis⁡θfor every real θ.0 = 0\operatorname{cis}\theta \quad \text{for every real } \theta.

Every angle fits equally well, so no angle is singled out, and arg⁡0\arg 0 and Arg⁡0\operatorname{Arg} 0 are left undefined. The polar form rcis⁡θr\operatorname{cis}\theta with r>0r > 0 applies only to nonzero numbers.

In plain words The arrow to 00 has length zero, so it points nowhere in particular. It is the same situation as the pole in polar coordinates.

Example Both 0cis⁡00\operatorname{cis}0 and 0cis⁡π20\operatorname{cis}\frac{\pi}{2} equal 00, so the angle of zero cannot be recovered from the number. That is why statements about arguments, such as the rule for dividing in polar form, assume the numbers involved are not zero.

Rectangular conversion

Definition To convert z=rcis⁡θz = r\operatorname{cis}\theta to rectangular form, use

a=rcos⁡θ,b=rsin⁡θ,z=a+bi.\begin{aligned} a &= r\cos\theta, \\[0.4em] b &= r\sin\theta, \\[0.4em] z &= a + bi. \end{aligned}

To convert a nonzero z=a+biz = a + bi to polar form, use

r=a2+b2,θ=atan2⁡(b,a),\begin{aligned} r &= \sqrt{a^2 + b^2}, \\[0.4em] \theta &= \operatorname{atan2}(b, a), \end{aligned}

which gives the principal argument. Any angle coterminal with it also works.

In plain words This is the same as converting a point between polar and rectangular coordinates, with aa and bb in place of xx and yy. The quadrant of (a,b)(a, b) decides the angle, not the tangent alone.

Example Going to rectangular form,

6cis⁡4π3=6(−12)+6(−32)i=−3−33 i.\begin{aligned} 6\operatorname{cis}\frac{4\pi}{3} &= 6\left(-\frac{1}{2}\right) + 6\left(-\frac{\sqrt{3}}{2}\right)i \\[0.8em] &= -3 - 3\sqrt{3}\,i. \end{aligned}

Going to polar form, the number z=−3+4iz = -3 + 4i has r=9+16=5r = \sqrt{9 + 16} = 5. Its angle is not a special one, and since a<0a < 0 and b>0b > 0,

Arg⁡z=arctan⁡(−43)+π=π−arctan⁡43≈2.214,\begin{aligned} \operatorname{Arg} z &= \arctan\left(-\frac{4}{3}\right) + \pi \\[0.8em] &= \pi - \arctan\frac{4}{3} \\[0.8em] &\approx 2.214, \end{aligned}

so −3+4i≈5cis⁡2.214-3 + 4i \approx 5\operatorname{cis} 2.214.

Check yourself

  1. Write −3+3i-3 + 3i in polar form with its principal argument.
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The modulus is 9+9=32\sqrt{9 + 9} = 3\sqrt{2}. The point is in the second quadrant on the line y=−xy = -x, so the principal argument is 3π4\frac{3\pi}{4} and

−3+3i=32cis⁡3π4.-3 + 3i = 3\sqrt{2}\operatorname{cis}\frac{3\pi}{4}.
  1. Write 5cis⁡(−2π3)5\operatorname{cis}(-\frac{2\pi}{3}) in rectangular form.
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Multiplying out the cosine and sine gives

5cos⁡(−2π3)+5isin⁡(−2π3)=−52−532 i.\begin{aligned} &5\cos\left(-\frac{2\pi}{3}\right) + 5i\sin\left(-\frac{2\pi}{3}\right) \\[0.8em] &\quad = -\frac{5}{2} - \frac{5\sqrt{3}}{2}\,i. \end{aligned}
  1. Write −4-4 in polar form, list all of its arguments and give its principal argument.
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The modulus is 44 and the point lies on the negative real axis, so −4=4cis⁡π-4 = 4\operatorname{cis}\pi. Its arguments are π+2kπ\pi + 2k\pi for integers kk, which are the odd multiples of π\pi. The principal argument is π\pi, not −π-\pi, because the interval (−π,π](-\pi, \pi] contains π\pi and leaves out −π-\pi.

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