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Topos Uranos
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The polar form of complex numbers

Summary

This lesson writes every nonzero complex number by means of its modulus and an argument, the polar angle of the point that represents it, and so obtains its polar or trigonometric form, z=r(cos⁡θ+isin⁡θ)z = r\left( \cos \theta + i \sin \theta \right). It describes all the arguments of a complex number, defines the principal argument and computes it with the arctangent corrected according to the quadrant, and proves when two polar forms represent the same number. With the sum and difference formulas it proves the central theorem: when two complex numbers are multiplied their moduli are multiplied and their arguments are added, and when they are divided the moduli are divided and the arguments are subtracted; from it follow the argument of the conjugate and that of the inverse, and it is pointed out that the principal argument of a product is the sum of the principal arguments only up to a full turn. Finally, it interprets the product by a complex number as a rotation composed with a dilation, which generalizes the quarter turn of the product by ii, and proves that the triangles with vertices 00, 11, ww and 00, zz, zwzw are similar.

Learning objectives

  1. Write a nonzero complex number in polar form, compute its principal argument according to the quadrant and decide when two polar forms represent the same number.
  2. Prove with the sum and difference formulas that in a product the moduli are multiplied and the arguments are added, and deduce the rules for the quotient, the conjugate and the inverse.
  3. Interpret the product by a complex number as a rotation composed with a dilation, and prove the similarity of the triangles with vertices 00, 11, ww and 00, zz, zwzw.
  4. Apply the polar form to exact values of trigonometric ratios, to identities between arctangents, to equations involving zz and its conjugate and to loci defined by arguments.

The polar form of a complex number

In the lesson of the course Algebra and Geometry I on the complex numbers, each complex number z=a+biz = a + bi was represented as the point (a,b)(a, b) of the plane, its modulus ∣z∣\abs{z} was defined as the distance from that point to the origin, and it was proved that the product by ii rotates each point through a quarter turn. The product by an arbitrary complex number was left pending, since its effect cannot be read in rectangular form: the real part of (a+bi)(c+di)(a + bi)(c + di) is ac−bdac - bd, an expression that reveals no geometry. Now, a point of the plane can also be named by its distance to the origin and by the direction in which it is seen from there, as in the lesson on polar coordinates. If this is so for points, it is so for complex numbers; and in that second description, as will be proved, the product becomes transparent.

Let z=a+biz = a + bi be a nonzero complex number. An argument of zz is any real number θ\theta such that a=∣z∣cos⁡θa = \abs{z} \cos \theta and b=∣z∣sin⁡θb = \abs{z} \sin \theta. If r=∣z∣r = \abs{z} and θ\theta is an argument of zz, the following expression is a polar form, or trigonometric form, of zz.
z=r(cos⁡θ+isin⁡θ)z = r\left( \cos \theta + i \sin \theta \right)

Every nonzero complex number has arguments. Indeed, by the theorem on the passage from Cartesian to polar coordinates, proved in the lesson on polar coordinates, the point (a,b)(a, b), distinct from the pole, has a polar representation (r,θ)(r, \theta) with r>0r > 0, and its radius is r=a2+b2r = \sqrt{a^2 + b^2}, that is, the modulus of zz; the equalities a=rcos⁡θa = r \cos \theta and b=rsin⁡θb = r \sin \theta then say that the polar angle θ\theta is an argument of zz. An argument of zz is therefore the angle that the line segment joining the origin to zz makes with the positive real half-axis, measured in radians. Zero, on the other hand, has no argument: the equality 0=0(cos⁡θ+isin⁡θ)0 = 0\left( \cos \theta + i \sin \theta \right) holds for every θ\theta, and no angle singles it out. To designate any of the arguments of zz one writes arg⁡z\arg z, which is read “argument of zz”.

The following theorem says that the modulus and the argument, unlike polar coordinates with negative radius, determine the complex number without ambiguity, up to full turns. Every computation with polar forms depends on it: to prove that two complex numbers are equal it will suffice to compare their moduli and their arguments.

TheoremEquality in polar form

Let rr, r1r_1 and r2r_2 be positive reals, and θ\theta, θ1\theta_1 and θ2\theta_2 real numbers. The complex number r(cos⁡θ+isin⁡θ)r\left( \cos \theta + i \sin \theta \right) has modulus rr and argument θ\theta. Moreover, the complex numbers r1(cos⁡θ1+isin⁡θ1)r_1\left( \cos \theta_1 + i \sin \theta_1 \right) and r2(cos⁡θ2+isin⁡θ2)r_2\left( \cos \theta_2 + i \sin \theta_2 \right) are equal if and only if r1=r2r_1 = r_2 and θ1−θ2=2kπ\theta_1 - \theta_2 = 2k\pi for some integer kk. The proof uses the Pythagorean identity, the periodicity of sine and cosine and the characterization of coterminal angles from the lesson on angles and their measure: two angles give the same point of the unit circle exactly when they differ by an integer multiple of 2π2\pi.

Proof

  1. ∣z∣2=r2cos⁡2θ+r2sin⁡2θ=r2\abs{z}^2 = \resaltar{r^2 \cos^2 \theta + r^2 \sin^2 \theta} = \resaltar{r^2}

    Let z=r(cos⁡θ+isin⁡θ)z = r\left( \cos \theta + i \sin \theta \right), with r>0r > 0. Its real part is rcos⁡θr \cos \theta and its imaginary part, rsin⁡θr \sin \theta; by the Pythagorean identity, the square of its modulus is r2r^2.

  2. ∣z∣=r,rcos⁡θ=∣z∣cos⁡θ,rsin⁡θ=∣z∣sin⁡θ\dato{m}{\resaltar{\abs{z} = r}}, \qquad r \cos \theta = \resaltar{\abs{z} \cos \theta}, \qquad r \sin \theta = \abs{z} \sin \theta

    The numbers ∣z∣\abs{z} and rr are positive and have the same square; therefore they are equal, and the parts of zz are those the definition requires: θ\theta is an argument of zz. Every expression of this form with r>0r > 0 is therefore a polar form.

  3. r1(cos⁡θ1+isin⁡θ1)=r2(cos⁡θ2+isin⁡θ2)→r1=r2r_1 \left( \cos \theta_1 + i \sin \theta_1 \right) = r_2 \left( \cos \theta_2 + i \sin \theta_2 \right) \rightarrow \dato{r}{\resaltar{r_1 = r_2}}

    If the complex numbers of modulus r1r_1 and r2r_2 are equal, they have the same modulus, and by the preceding step their moduli are r1r_1 and r2r_2.

  4. cos⁡θ1=cos⁡θ2∧sin⁡θ1=sin⁡θ2↔θ1−θ2=2kπ\resaltar{\cos \theta_1 = \cos \theta_2 \wedge \sin \theta_1 = \sin \theta_2} \leftrightarrow \resaltar{\theta_1 - \theta_2 = 2k\pi}

    We equate the real parts and the imaginary parts and divide by r1=r2r_1 = r_2, which is not zero: the angles θ1\theta_1 and θ2\theta_2 give the same point of the unit circle, so they are coterminal.

  5. cos⁡(θ2+2kπ)=cos⁡θ2,sin⁡(θ2+2kπ)=sin⁡θ2\cos(\resaltar{\theta_2 + 2k\pi}) = \resaltar{\cos \theta_2}, \qquad \sin(\theta_2 + 2k\pi) = \sin \theta_2

    Conversely, if r1=r2r_1 = r_2 and θ1=θ2+2kπ\theta_1 = \theta_2 + 2k\pi, by the periodicity of sine and cosine the two expressions have the same parts, and the complex numbers are equal.

It follows that, if θ\theta is an argument of zz, the arguments of zz are exactly the numbers θ+2kπ\theta + 2k\pi, with kk an integer: two arguments of the same complex number always differ by full turns. The equality arg⁡z=θ\arg z = \theta therefore means that θ\theta is one of the arguments of zz, and does not fix the others. It should also be noted that the theorem requires r>0r > 0: the expression −2(cos⁡π3+isin⁡π3)-2\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) is not a polar form, because −2-2 is not a modulus; since −cos⁡θ=cos⁡(θ+π)-\cos \theta = \cos(\theta + \pi) and −sin⁡θ=sin⁡(θ+π)-\sin \theta = \sin(\theta + \pi), the same number is written in polar form as 2(cos⁡4π3+isin⁡4π3)2\left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right).

The principal argument

Among the arguments of a nonzero complex number there is exactly one in the interval (−π,π]\ioc{-\pi}{\pi}, because that interval has length 2π2\pi and contains a single representative of each family θ+2kπ\theta + 2k\pi. That argument is called the principal argument of zz, is written Arg⁡z\operatorname{Arg} z and is read “principal argument of zz”. By the theorem of the lesson on polar coordinates, with aa and bb in the role of xx and yy, it is computed with the arctangent from the lesson on the inverse trigonometric functions, which only sees the first and fourth quadrants and must be corrected in the other two:

Arg⁡z={arctan⁡baa>0arctan⁡ba+πa<0∧b≥0arctan⁡ba−πa<0∧b<0π2a=0∧b>0−π2a=0∧b<0\operatorname{Arg} z = \begin{cases} \arctan \frac{b}{a} & a > 0 \\ \arctan \frac{b}{a} + \pi & a < 0 \wedge b \geq 0 \\ \arctan \frac{b}{a} - \pi & a < 0 \wedge b < 0 \\ \frac{\pi}{2} & a = 0 \wedge b > 0 \\ -\frac{\pi}{2} & a = 0 \wedge b < 0 \end{cases}
ExampleThe polar form of a complex number in the second quadrant

We write the complex number z=−3+iz = -\sqrt{3} + i in polar form and describe all its arguments. The figure follows the computation in the complex plane.

Proof

  1. ∣z∣=(−3)2+12=4=2\abs{z} = \sqrt{\resaltar{\left( -\sqrt{3} \right)^2 + 1^2}} = \sqrt{4} = \dato{m}{\resaltar{2}}

    The modulus is the distance from the point (−3,1)\left( -\sqrt{3}, 1 \right) to the origin.

  2. arctan⁡1−3=arctan⁡(−33)=−π6\arctan \frac{1}{-\sqrt{3}} = \resaltar{\arctan\left( -\frac{\sqrt{3}}{3} \right)} = \dato{a}{\resaltar{-\frac{\pi}{6}}}

    The arctangent of the quotient of the imaginary part by the real part gives an angle in the fourth quadrant: it is the argument of the opposite point, 3−i\sqrt{3} - i, and not that of zz.

  3. Arg⁡z=−π6+π=5π6\operatorname{Arg} z = \resaltar{-\frac{\pi}{6} + \pi} = \dato{t}{\resaltar{\frac{5\pi}{6}}}

    Since the real part is negative and the imaginary part is positive, the point is in the second quadrant, and the formula adds π\pi.

  4. z=2(cos⁡5π6+isin⁡5π6)=2(−32+12i)=−3+iz = 2\left( \resaltar{\cos \frac{5\pi}{6}} + i \sin \frac{5\pi}{6} \right) = 2\left( \resaltar{-\frac{\sqrt{3}}{2}} + \frac{1}{2} i \right) = -\sqrt{3} + i

    We check the polar form with the values of the cosine and sine of 5π6\frac{5\pi}{6}, which are obtained from those of π6\frac{\pi}{6} by the symmetry of the supplementary angle.

  5. arg⁡z=5π6+2kπ,k∈Z\arg z = \resaltar{\frac{5\pi}{6} + 2k\pi}, \qquad k \in \mathbb{Z}

    The other arguments differ from the principal one by full turns: for example, −7π6-\frac{7\pi}{6} and 17π6\frac{17\pi}{6} are also arguments of zz, but they are not in (−π,π]\ioc{-\pi}{\pi}.

Complex numbers lying on the axes have polar forms that are worth recognizing at once. Positive reals have principal argument 00, and negative ones, π\pi; purely imaginary numbers bibi with b>0b > 0 have principal argument π2\frac{\pi}{2}, and with b<0b < 0, −π2-\frac{\pi}{2}. Thus, for example:

−1=cos⁡π+isin⁡π,i=cos⁡π2+isin⁡π2,−2i=2(cos⁡(−π2)+isin⁡(−π2))-1 = \cos \pi + i \sin \pi, \qquad i = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}, \qquad -2i = 2\left( \cos\left( -\frac{\pi}{2} \right) + i \sin\left( -\frac{\pi}{2} \right) \right)

The interval (−π,π]\ioc{-\pi}{\pi} is a convention, the same as in the lesson on polar coordinates; other texts use [0,2π)\ico{0}{2\pi}. What does not depend on the convention is the set of all arguments, and that is why the theorems of this lesson are stated with arbitrary arguments, unless expressly stated otherwise.