Complex Numbers IV: De Moivre's Theorem
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
Use De Moivre's theorem to simplify:
(a)\(\left(3\, ext{cis}\dfrac{2\pi}{5}
ight)^5\)
(b)\(\left( ext{cis}\dfrac{\pi}{3}
ight)^4\)
(c)\((\sqrt{3}+i)^6\)
(a)
Apply De Moivre
\(3^5\, ext{cis}(2\pi)=243 imes1\)
(b)
Apply De Moivre
\( ext{cis} frac{4\pi}{3}=\cos frac{4\pi}{3}+i\sin frac{4\pi}{3}\)
Answer
\(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\)
(c)
Convert to polar: \(r=2,\; heta= frac{\pi}{6}\)
\(2^6\, ext{cis}\pi=64(-1)\)
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
\(\sum_{k=0}^5inom{5}{k}i^k=1+5i-10-10i+5+i=(-4-4i)\)
Polar: \(1+i=\sqrt2\, ext{cis} frac{\pi}{4}\)
\((\sqrt2)^5\, ext{cis} frac{5\pi}{4}=4\sqrt2\!\left(- frac{\sqrt2}{2}- frac{\sqrt2}{2}i
ight)=-4-4i\) ✓
Question 3
Convert \(1-i\sqrt{3}\) to polar form. Hence find \(\dfrac{1}{(1-i\sqrt3)^4}\).
Polar
\(r=2;\; heta=- frac{\pi}{3}\); so \(z=2\, ext{cis}\!\left(- frac{\pi}{3}
ight)\)
Raise to power 4
\(z^4=16\, ext{cis}\!\left(- frac{4\pi}{3}
ight)\); reciprocal: \( frac{1}{16}\, ext{cis} frac{4\pi}{3}\)
Answer
\(\dfrac{1}{16}\!\left(-\dfrac{1}{2}-\dfrac{\sqrt3}{2}i
ight)=-\dfrac{1}{32}-\dfrac{\sqrt3}{32}i\)
Question 4
For complex numbers \(w=2+2i\) and \(z= ext{cis}\!\left(\dfrac{3\pi}{4}
ight)\), find \(w^2z^3\) in Cartesian form.
Convert \(w\) to polar
\(w=2\sqrt2\, ext{cis} frac{\pi}{4}\Rightarrow w^2=8\, ext{cis} frac{\pi}{2}=8i\)
\(z^3= ext{cis} frac{9\pi}{4}= ext{cis} frac{\pi}{4}= frac{\sqrt2}{2}+ frac{\sqrt2}{2}i\)
\(w^2z^3=8i\!\left( frac{\sqrt2}{2}+ frac{\sqrt2}{2}i
ight)=4\sqrt2 i-4\sqrt2\)
Answer
\(-4\sqrt{2}+4\sqrt{2}\,i\)
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