Complex Numbers IV: De Moivre's Theorem
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
Use De Moivre's theorem to simplify:
(a)\(\left(3\,\text{cis}\dfrac{2\pi}{5}\right)^5\)
(b)\(\left(\text{cis}\dfrac{\pi}{3}\right)^4\)
(c)\((\sqrt{3}+i)^6\)
(a)
Apply De Moivre
\(3^5\,\text{cis}(2\pi)=243\times1\)
(b)
Apply De Moivre
\(\text{cis}\tfrac{4\pi}{3}=\cos\tfrac{4\pi}{3}+i\sin\tfrac{4\pi}{3}\)
Answer
\(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\)
(c)
Convert to polar: \(r=2,\;\theta=\tfrac{\pi}{6}\)
\(2^6\,\text{cis}\pi=64(-1)\)
Question 2
Expand \((1+i)^5\) using the binomial expansion. Convert \(1+i\) to polar form and hence verify your answer using De Moivre's theorem.
Binomial
\(\sum_{k=0}^5\binom{5}{k}i^k=1+5i-10-10i+5+i=(-4-4i)\)
Polar: \(1+i=\sqrt2\,\text{cis}\tfrac{\pi}{4}\)
\((\sqrt2)^5\,\text{cis}\tfrac{5\pi}{4}=4\sqrt2\!\left(-\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}i\right)=-4-4i\) ✓
Question 3
Convert \(1-i\sqrt{3}\) to polar form. Hence find \(\dfrac{1}{(1-i\sqrt3)^4}\).
Polar
\(r=2;\;\theta=-\tfrac{\pi}{3}\); so \(z=2\,\text{cis}\!\left(-\tfrac{\pi}{3}\right)\)
Raise to power 4
\(z^4=16\,\text{cis}\!\left(-\tfrac{4\pi}{3}\right)\); reciprocal: \(\tfrac{1}{16}\,\text{cis}\tfrac{4\pi}{3}\)
Answer
\(\dfrac{1}{16}\!\left(-\dfrac{1}{2}-\dfrac{\sqrt3}{2}i\right)=-\dfrac{1}{32}-\dfrac{\sqrt3}{32}i\)
Question 4
For complex numbers \(w=2+2i\) and \(z=\text{cis}\!\left(\dfrac{3\pi}{4}\right)\), find \(w^2z^3\) in Cartesian form.
Convert \(w\) to polar
\(w=2\sqrt2\,\text{cis}\tfrac{\pi}{4}\Rightarrow w^2=8\,\text{cis}\tfrac{\pi}{2}=8i\)
\(z^3=\text{cis}\tfrac{9\pi}{4}=\text{cis}\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}i\)
\(w^2z^3=8i\!\left(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}i\right)=4\sqrt2 i-4\sqrt2\)
Answer
\(-4\sqrt{2}+4\sqrt{2}\,i\)
Generated by Mathski · mathski.io · IB Mathematics AA Shadow Worksheets