This note covers three linked results for complex numbers. Roots of polynomials with real coefficients come in conjugate pairs, De Moivre's theorem gives a fast way to raise a complex number to a power, and the same theorem lets you find all the roots of a complex number. Together they turn long calculations into short ones.
Key ideas
Conjugate roots. Suppose a polynomial has only real coefficients. If a complex number \(p+qi\) is a root, then its conjugate \(p-qi\) is also a root. So non-real roots always come in pairs. This is why a cubic with real coefficients must have at least one real root, since three roots cannot all be paired up. A quadratic factor made from a conjugate pair always has real coefficients:
\[\bigl(z-(p+qi)\bigr)\bigl(z-(p-qi)\bigr)=z^2-2pz+(p^2+q^2).\]
So the pair has sum \(2p\) and product \(p^2+q^2\). This is how you build a polynomial from known roots, or find the remaining roots when one is given.
De Moivre's theorem. For any integer \(n\),
\[\bigl(r(\cos\theta+i\sin\theta)\bigr)^n=r^n\bigl(\cos n\theta+i\sin n\theta\bigr),\]
or in Euler form \((re^{i\theta})^n=r^ne^{in\theta}\). The modulus is raised to the power \(n\) and the argument is multiplied by \(n\). To use it, first convert the number to polar form. The theorem also works for rational powers such as \(\frac1n\), but then there are several answers, as described below.
Roots of a complex number. To solve \(z^n=w\), write \(w=Re^{i\phi}\). Because adding \(2\pi k\) to the argument gives the same number, \(w=Re^{i(\phi+2\pi k)}\) for any integer \(k\). Taking the \(n\)th root gives
\[z=R^{1/n}\,e^{\,i(\phi+2\pi k)/n},\qquad k=0,1,\dots,n-1.\]
Values of \(k\) beyond \(n-1\) just repeat the same roots. There are exactly \(n\) distinct roots. They all have the same modulus \(R^{1/n}\), so on the Argand diagram they lie on a circle, and the angle between neighbouring roots is \(\frac{2\pi}n\). They form the corners of a regular \(n\)-sided polygon.
The solutions of \(z^n=1\) are called the \(n\)th roots of unity. They are \(e^{2\pi ik/n}\).
Using the results well
De Moivre's theorem is repeated multiplication in disguise. Multiplying \(r\,\mathrm{cis}\,\theta\) by itself \(n\) times multiplies the modulus \(n\) times and adds the argument \(n\) times. It also works for negative integers: \((r\,\mathrm{cis}\,\theta)^{-1}=\frac1r\,\mathrm{cis}(-\theta)\).
For polynomial questions, remember what the conjugate root theorem tells you about counting. A cubic with real coefficients has either three real roots, or one real root and one conjugate pair. A quartic has four real roots, two real roots and a pair, or two pairs. If you are told that a quartic with real coefficients has a root \(1+i\), you immediately know \(1-i\) is a root too, so \(z^2-2z+2\) is a factor, and you can divide it out to find the rest.
When solving \(z^n=w\), you can find the first root and then get the others by adding \(\frac{2\pi}n\) to the argument each time. It is a quick way to check you have not made an arithmetic slip, because after \(n\) steps you should return to the start.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(p+qi\) a root \(\Rightarrow\) \(p-qi\) a root | Conjugate root theorem. Needs a polynomial with real coefficients. |
| \(z^2-2pz+(p^2+q^2)\) | Real quadratic factor with roots \(p\pm qi\). |
| \((r\,\mathrm{cis}\,\theta)^n=r^n\,\mathrm{cis}\,(n\theta)\) | De Moivre's theorem, for integer \(n\). Use for powers of complex numbers. |
| \(z^n=Re^{i\phi}\Rightarrow z=R^{1/n}e^{i(\phi+2\pi k)/n}\) | The \(n\) roots, with \(k=0,1,\dots,n-1\). |
| Roots of unity | The solutions of \(z^n=1\). They are equally spaced on the unit circle. |
Worked examples
Example 1 A cubic polynomial with real coefficients and leading coefficient 1 has roots \(2\) and \(1+2i\). Find the polynomial.
The coefficients are real, so \(1-2i\) is also a root. The three roots are \(2\), \(1+2i\) and \(1-2i\).
The conjugate pair gives \(z^2-2(1)z+(1^2+2^2)=z^2-2z+5\).
Multiply by the factor for the real root: \((z-2)(z^2-2z+5)=z^3-2z^2+5z-2z^2+4z-10\).
Collect terms: \(z^3-4z^2+9z-10\).
Check: at \(z=2\), \(8-16+18-10=0\).
Answer: \(z^3-4z^2+9z-10\)
Example 2 Find \((\sqrt3-i)^5\) in the form \(a+bi\).
Modulus: \(r=\sqrt{3+1}=2\). The point \((\sqrt3,-1)\) is in the fourth quadrant, with reference angle \(\tan^{-1}\frac1{\sqrt3}=\frac\pi6\), so \(\theta=-\frac\pi6\).
Apply De Moivre: \((\sqrt3-i)^5=2^5\left(\cos\left(-\tfrac{5\pi}6\right)+i\sin\left(-\tfrac{5\pi}6\right)\right)\).
Evaluate: \(\cos\left(-\tfrac{5\pi}6\right)=-\tfrac{\sqrt3}2\) and \(\sin\left(-\tfrac{5\pi}6\right)=-\tfrac12\).
So the result is \(32\left(-\tfrac{\sqrt3}2-\tfrac12i\right)=-16\sqrt3-16i\).
Answer: \(-16\sqrt3-16i\)
Example 3 Solve \(z^3=-8\), giving the roots in the form \(a+bi\).
Write \(-8\) in polar form: modulus 8, argument \(\pi\). So \(-8=8e^{i(\pi+2\pi k)}\).
Take the cube root: \(z=8^{1/3}e^{i(\pi+2\pi k)/3}=2e^{i(\pi+2\pi k)/3}\).
\(k=0\): \(z=2e^{i\pi/3}=2\left(\tfrac12+\tfrac{\sqrt3}2i\right)=1+\sqrt3\,i\).
\(k=1\): \(z=2e^{i\pi}=-2\).
\(k=2\): \(z=2e^{5i\pi/3}=2\left(\tfrac12-\tfrac{\sqrt3}2i\right)=1-\sqrt3\,i\).
The roots are \(\frac{2\pi}3\) apart on a circle of radius 2, and the complex ones are conjugates, as the polynomial \(z^3+8\) has real coefficients.
Answer: \(z=-2\), \(z=1+\sqrt3\,i\) or \(z=1-\sqrt3\,i\)
Common errors
- Assuming conjugate roots when the polynomial has a non-real coefficient. The theorem needs all coefficients to be real.
- Applying De Moivre to a number in Cartesian form. Convert to polar form first.
- Forgetting to raise the modulus to the power. In \((2\,\mathrm{cis}\,\theta)^5\) the modulus becomes 32, not 2.
- Finding only one root of \(z^n=w\). There are \(n\) roots, so list \(k=0\) to \(n-1\).
- Dividing only \(\phi\) by \(n\) and forgetting the \(2\pi k\). The \(2\pi k\) must be added before dividing, otherwise all the roots are missed except the first.
Exam tip: After finding the roots, check that they are equally spaced and have the same modulus. If the polynomial has real coefficients, also check that the non-real roots form conjugate pairs.
Check yourself
- A quadratic with real coefficients has a root \(2-3i\). State the other root and find the quadratic with leading coefficient 1.
Answer
The other root is \(2+3i\). The sum is 4 and the product is \(4+9=13\), so the quadratic is \(z^2-4z+13\). - Simplify \(\left(\cos\frac\pi6+i\sin\frac\pi6\right)^{12}\).
Answer
By De Moivre the angle becomes \(12\times\frac\pi6=2\pi\). So the result is \(\cos2\pi+i\sin2\pi=1\). - Find \((1+i)^6\) in the form \(a+bi\).
Answer
\(1+i=\sqrt2\,\mathrm{cis}\frac\pi4\). Then \((1+i)^6=(\sqrt2)^6\,\mathrm{cis}\frac{6\pi}4=8\,\mathrm{cis}\frac{3\pi}2=8(0-i)=-8i\). - Find all solutions of \(z^4=16\).
Answer
\(16=16e^{2\pi ik}\), so \(z=2e^{2\pi ik/4}=2e^{i\pi k/2}\) for \(k=0,1,2,3\). The solutions are \(2\), \(2i\), \(-2\) and \(-2i\).