[ALGEBRA] Polar form

This page describes a complex number by its distance from the origin and its direction, instead of by its Cartesian components. In these polar coordinates on the Argand plane, multiplication multiplies moduli and adds arguments, which makes powers and roots straightforward.

For $z=x+iy\ne0$ with polar coordinates $(r,\theta)$ ,

r=|z|,\qquad x=r\cos\theta,\qquad y=r\sin\theta.

Argument and polar form

[DEFINITION]

Complex number

A complex number is a number of the form

z=a+bi,\qquad a,b\in\mathbb R,\qquad i^2=-1.

The expression $a+bi$ is the Cartesian form of $z$ . Its real part is $\operatorname{Re}(z)=a$ and its imaginary part is $\operatorname{Im}(z)=b$ .

[1]:Argand plane
[2]:Complex conjugate
[3]:Complex modulus
[4]:Add, multiply and divide complex numbers
7 more

[DEFINITION]

Complex modulus

The modulus of $z=a+bi$ is

|z|=\sqrt{a^2+b^2}.

[DEFINITION]

Argument

For $z\ne0$ , an argument of $z$ is a real number $\theta$ satisfying

\cos\theta=\frac{\operatorname{Re}(z)}{|z|}, \qquad \sin\theta=\frac{\operatorname{Im}(z)}{|z|}.

Arguments are determined up to addition of $2\pi k$ , where $k\in\mathbb Z$ . The principal argument $\operatorname{Arg}(z)$ is the unique argument in

-\pi<\operatorname{Arg}(z)\le\pi.

Choose the argument from the point's quadrant, not from $\tan\theta$ alone: tangent has period $\pi$ , so it cannot distinguish opposite quadrants.

The principal argument is one conventional choice; every argument remains valid. The number $z=0$ has no argument, because the origin does not determine a direction.

[1]:Complex number
[2]:Complex modulus
[3]:Polar form
[4]:Principal arguments of products and quotients
[5]:Read loci in the Argand plane

[DEFINITION]

Polar form

For $z\ne0$ , the polar form of $z$ is

z=r(\cos\theta+i\sin\theta)=re^{i\theta}, \qquad r=|z|>0,

where $\theta$ is an argument of $z$ . The expressions $a+bi$ and $re^{i\theta}$ are two writings of the same complex number.

Euler's formula,

e^{i\theta}=\cos\theta+i\sin\theta,

identifies the two expressions in the statement. Replacing $\theta$ by $\theta+2\pi k$ leaves $z$ unchanged.

Euler's formula is taken as given here rather than derived. The Taylor series of the exponential, sine and cosine are the usual route to a proof.

Multiplication by $e^{i\theta}$ rotates a point through angle $\theta$ without changing its modulus; multiplication by $re^{i\theta}$ also scales it by $r$ . In coordinates, for $z=x+iy$ ,

\begin{pmatrix} \operatorname{Re}(e^{i\theta}z)\\ \operatorname{Im}(e^{i\theta}z) \end{pmatrix} = \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}

[EXAMPLE]

Convert between the two forms

For $z=1+i$ ,

|z|=\sqrt{1^2+1^2}=\sqrt2.

The point lies in the first quadrant and $\tan\theta=1$ , so $\theta=\pi/4$ is a principal argument. Hence

1+i=\sqrt2\,e^{i\pi/4}.

Conversely,

\begin{aligned} 2e^{-3\pi i/4} &=2\left(\cos\left(-\frac{3\pi}{4}\right) +i\sin\left(-\frac{3\pi}{4}\right)\right)\\ &=-\sqrt2-\sqrt2\,i. \end{aligned}

[routine]

[1]:Polar form

[THEOREM]

Modulus laws

For $z,z_1,z_2\in\mathbb C$ with $z_2\ne0$ ,

z\overline z=|z|^2,\qquad |z_1z_2|=|z_1||z_2|,\qquad \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|},

|\operatorname{Re}(z)|\le|z|, \qquad |\operatorname{Im}(z)|\le|z|, \qquad |z_1+z_2|\le|z_1|+|z_2|,

|z_1-z_2|\ge\bigl||z_1|-|z_2|\bigr|, \qquad \left|\sum_{j=1}^N z_j\right| \le\sum_{j=1}^N|z_j|\quad(N\ge1).

The inequality for $z_1+z_2$ is the triangle inequality in the Argand plane; the finite-sum version follows by applying it repeatedly. The reverse triangle inequality follows from

|z_1|=|(z_1-z_2)+z_2|\le |z_1-z_2|+|z_2|

and the same argument with $z_1$ and $z_2$ interchanged.

[1]:Complex modulus

Multiplication and division in polar form

[METHOD]

Multiply and divide in polar form

If $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ , then

z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}

and, for $z_2\ne0$ ,

\frac{z_1}{z_2} =\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}.

In particular,

(re^{i\theta})^{-1}=\frac1r e^{-i\theta}.

Geometrically, a product multiplies the moduli and adds the arguments and a quotient divides the moduli and reverses the second rotation.

To see why the angles add, multiply $(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)$ and use the angle-addition identities: its real and imaginary parts become $\cos(\theta_1+\theta_2)$ and $\sin(\theta_1+\theta_2)$ . The quotient rule then follows by multiplying by the inverse.

[EXAMPLE]

Perform product and quotient calculations

Let

z_1=\sqrt2\,e^{3\pi i/2}, \qquad z_2=\pi e^{3\pi i/2}.

Both factors point down the imaginary axis. Their arguments add to $3\pi$ , so the product points along the negative real axis:

z_1z_2=\pi\sqrt2\,e^{3\pi i}= -\pi\sqrt2.

If instead $z_2=\pi e^{3\pi i/4}$ , then

\frac{z_1}{z_2} =\frac{\sqrt2}{\pi}e^{i(3\pi/2-3\pi/4)} =\frac{\sqrt2}{\pi}e^{3\pi i/4}.

[routine]

[1]:Multiply and divide in polar form

[PROPOSITION]

Principal arguments of products and quotients

For non-zero $z_1,z_2\in\mathbb C$ , there are integers $k,\ell\in\mathbb Z$ such that

\operatorname{Arg}(z_1z_2) =\operatorname{Arg}(z_1)+\operatorname{Arg}(z_2)+2\pi k,

\operatorname{Arg}\left(\frac{z_1}{z_2}\right) =\operatorname{Arg}(z_1)-\operatorname{Arg}(z_2)+2\pi\ell.

Adding or subtracting principal arguments gives an argument of the result, but it may lie outside the principal interval $(-\pi,\pi]$ . The integers $k$ and $\ell$ record the reduction back into that interval. For instance, $\operatorname{Arg}((-1)(-1))=0$ , whereas $\operatorname{Arg}(-1)+\operatorname{Arg}(-1)=2\pi$ ; the latter is an argument of the product, but not its principal argument.

Two polar forms describe the same non-zero number exactly when their moduli agree and their arguments differ by an integer multiple of $2\pi$ .

[1]:Argument
[2]:Multiply and divide in polar form

Powers through De Moivre's formula

[THEOREM]

De Moivre's formula

For $\theta\in\mathbb R$ and $n\in\mathbb Z$ ,

\left(\cos\theta+i\sin\theta\right)^n =\cos(n\theta)+i\sin(n\theta).

More generally, if $z=re^{i\theta}$ with $r>0$ , then

z^n=r^ne^{in\theta}.

For positive integers, use induction. The case $n=1$ is immediate and if the formula holds at $n=k$ , multiplying by $re^{i\theta}$ gives

z^{k+1}=r^ke^{ik\theta}re^{i\theta}=r^{k+1}e^{i(k+1)\theta}.

The zero case is $z^0=1$ and the negative-integer case follows from the multiplicative inverse. The condition $r>0$ is essential for negative powers, because $0$ has no multiplicative inverse.

[EXAMPLE]

Calculate a power

For $z=\sqrt2\,e^{3\pi i/4}$ ,

z^{104} =(\sqrt2)^{104}e^{104(3\pi i/4)} =2^{52}e^{78\pi i} =2^{52}.

Polar form handles the large exponent by treating the modulus and the argument separately.

[routine]

[1]:De Moivre's formula

Roots of complex numbers

[DEFINITION]

An nth root of a complex number

For a complex number $z_0$ and a positive integer $n$ , an $n$ th root is a complex number $z$ satisfying

z^n=z_0.

[THEOREM]

Find the nth roots

If $z_0\ne0$ , $n$ is a positive integer and $z_0=r_0e^{i\theta_0}$ , its distinct $n$ th roots are

z_k=\sqrt[n]{r_0}\exp\left(i\frac{\theta_0+2\pi k}{n}\right), \qquad k=0,1,\ldots,n-1.

They lie on a circle of radius $\sqrt[n]{r_0}$ centred at the origin and are equally spaced by an angle of $2\pi/n$ . A non-zero complex number has exactly $n$ distinct $n$ th roots.

To derive the formula, write $z=\rho e^{i\phi}$ . If $z^n=z_0$ , equality of moduli gives $\rho^n=r_0$ , while equality of arguments gives $n\phi=\theta_0+2\pi k$ . Taking $k=0,\ldots,n-1$ produces all distinct possibilities: two listed angles differ by less than $2\pi$ , so they cannot describe the same point. Replacing $k$ by $k+n$ adds $2\pi$ to the root's angle, so any other integer $k$ repeats one of them. If $z_0=0$ , the only $n$ th root is $z=0$ .

[EXAMPLE]

Find the eighth roots of unity

Since $1=e^{2\pi i k}$ for every integer $k$ , the eighth roots of unity are

z_k=e^{2\pi ik/8},\qquad k=0,1,\ldots,7.

They are the eight equally spaced points on the unit circle. For example,

z_0=1,\qquad z_1=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\qquad z_2=i,

and the remaining roots continue around the circle.

[routine]

[1]:Find the nth roots

The same polar multiplication gives the trigonometric identities in trigonometric and geometric applications. An $n$ th root solves $z^n=z_0$ ; polynomial roots extend the question to zeros of general polynomials.

Polar form

Complex number

Complex modulus

On this page