Polar Form And Powers Of Complex Numbers

Modulus and Argument

Complex multiplication combines changes in size with changes in angle. Polar form makes both visible: the modulus records size, while an argument records direction in the complex plane. This representation turns repeated multiplication into a power of a real number and multiplication of an angle, and it also provides a systematic way to find complex roots.

Write a complex number as \(z=x+iy\), where \(x,y\in\mathbb{R}\) and \(i^2=-1\). Its modulus is \(|z|=\sqrt{x^2+y^2}\). If \(z\ne0\), then \(|z|>0\), and the point \((x/|z|,y/|z|)\) lies on the unit circle. An angle \(\theta\) describing that point satisfies \(\cos\theta=x/|z|\) and \(\sin\theta=y/|z|\).

An argument is not usually unique: angles differing by a whole turn describe the same direction. When a single angle is needed, one may choose a representative from a fixed interval, but calculations with roots must account for all angles differing by multiples of \(2\pi\). For a nonzero complex number, polar form exists because its normalized coordinates lie on the unit circle; the cosine and sine of some angle give those coordinates.

Multiplication and De Moivre’s Formula

When two nonzero complex numbers are in polar form, multiplying them multiplies their moduli and adds their angles. This follows by expanding the product and using the sine and cosine addition identities:

Repeated multiplication therefore gives a power rule. The angle in the result is understood modulo \(2\pi\), since adding a full turn does not change the complex number.

Proof. First let \(n\) be a positive integer. For \(n=1\), the formula is the given polar form. Suppose it holds for a positive integer \(n\). Multiplying the expression for \(z^n\) by \(z\), the multiplication identity above gives $$ z^{n+1}=r^{n+1}\bigl(\cos((n+1)\theta)+i\sin((n+1)\theta)\bigr). $$ Induction proves the formula for all positive integers. For \(n=0\), the right-hand side is \(1(\cos 0+i\sin 0)=1=z^0\). For a negative integer \(n=-m\), where \(m>0\), use \(z^{-m}=1/z^m\). Since \(\cos^2(m\theta)+\sin^2(m\theta)=1\), the reciprocal of \(\cos(m\theta)+i\sin(m\theta)\) is \(\cos(m\theta)-i\sin(m\theta)=\cos(-m\theta)+i\sin(-m\theta)\). Also \(r^{-m}=1/r^m\). These facts give the stated formula for \(n=-m\). \(\square\)

Finding Square Roots

A square root of \(z\) is a complex number \(w\) satisfying \(w^2=z\). For \(z\ne0\), polar form shows why there are two square roots: the modulus of \(w\) must be \(\sqrt{|z|}\), while twice an argument of \(w\) must be an argument of \(z\), up to a whole turn. Thus the two angles in one full turn are separated by \(\pi\), giving opposite roots. For \(z=0\), the only square root is \(0\).

This coordinate method works even when an angle is inconvenient to name exactly. For a nonzero number written in polar form, the same square roots can be obtained by taking the positive square root of its modulus and halving an argument, then including the angle that differs by \(\pi\). The coordinate check above confirms both solutions directly without relying on a particular choice of angle.

All the nth Roots

For a positive integer \(n\), an nth root of \(z\) is a complex number \(w\) such that \(w^n=z\). When \(z\ne0\), every root has positive modulus and therefore has an argument. If \(z=0\), its only nth root is \(0\): indeed, \(|w^n|=|w|^n=0\) forces \(|w|=0\), hence \(w=0\). The polar formula for the nonzero case is as follows.

Proof. First suppose \(z\ne0\). For each \(k=0,\dots,n-1\), apply De Moivre’s formula to \(w_k\). Its modulus becomes \(r\), and its angle becomes \(\theta+2k\pi\). Since adding \(2k\pi\) does not change cosine or sine, \(w_k^n=z\). The listed roots are distinct: if \(w_j=w_k\), their angles differ by an integer multiple of \(2\pi\), so \(2(j-k)\pi/n=2q\pi\) for some \(q\in\mathbb{Z}\). Hence \(j-k=qn\). Since \(j,k\) both lie between \(0\) and \(n-1\), this forces \(j=k\). Now let \(w\) be any nth root of \(z\). Write \(w=\rho(\cos\phi+i\sin\phi)\), where \(\rho>0\). Taking moduli in \(w^n=z\) gives \(\rho^n=r\), so \(\rho=\sqrt[n]{r}\). By De Moivre’s formula, the angle \(n\phi\) must be an argument of \(z\). Therefore \(n\phi=\theta+2m\pi\) for some \(m\in\mathbb{Z}\), and $$ \phi=\frac{\theta+2m\pi}{n}. $$ Write \(m=qn+k\) with \(q\in\mathbb{Z}\) and \(k\in\{0,\dots,n-1\}\). Then \(\phi=(\theta+2k\pi)/n+2q\pi\), which gives the same complex number as \(w_k\). Thus every root is on the list, and the earlier argument shows the listed roots are distinct. For \(z=0\), the modulus argument at the start gives the sole root \(w=0\). \(\square\)

Check: Using De Moivre’s formula, each number has modulus \(2\), so its cube has modulus \(8\). Their tripled angles are respectively \(\pi/2\), \(5\pi/2\), and \(9\pi/2\); each angle differs from \(\pi/2\) by a multiple of \(2\pi\). Thus each cube is \(8i\). For a direct check on the first, \((\sqrt{3}+i)^2=2+2\sqrt{3}i\), and \((2+2\sqrt{3}i)(\sqrt{3}+i)=2\sqrt{3}+2i+6i-2\sqrt{3}=8i\). The angle calculation checks the other two by the same power formula.

Common Pitfall: Omitting the Other Root Angles

A frequent error is to divide one argument by \(n\) and report only that single root. For example, dividing the argument \(\pi/2\) of \(8i\) by \(3\) gives the root \(\sqrt{3}+i\), but it does not give all cube roots. The missing angles arise because an argument may first be increased by any integer multiple of \(2\pi\). The complete set uses \((\pi/2+2k\pi)/3\) for \(k=0,1,2\), producing \(\sqrt{3}+i\), \(-\sqrt{3}+i\), and \(-2i\), each of which cubes to \(8i\).

The zero case also needs separate attention. The formula for nonzero roots uses an argument, and \(0\) has none. For \(z=0\), do not assign it an angle: use \(|w|^n=|w^n|=0\) to conclude directly that the only root is \(w=0\).

Polar form is useful whenever multiplication or repeated powers are involved, because it converts those operations into arithmetic on moduli and angles. It is especially effective for finding roots, but only if all angle choices are included and the zero case is handled separately.