Complex Numbers VI: Mixed
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
If \(z=2(\cos\theta+i\sin\theta)\), find \(z+\bar{z}\) and \(z\bar{z}\).
Conjugate
\(\bar{z}=2(\cos\theta-i\sin\theta)\)
Answer
\(z+\bar{z}=4\cos\theta\); \(z\bar{z}=|z|^2=4\)
Question 2
Show that if \(z=r\,\text{cis}\,\theta\) then \(\dfrac{1}{z}=\dfrac{1}{r}\text{cis}(-\theta)\). Hence find \(\dfrac{1}{1+i}\) without rationalising.
Polar of \(1+i\)
\(r=\sqrt2,\;\theta=\tfrac{\pi}{4}\Rightarrow\dfrac{1}{1+i}=\dfrac{1}{\sqrt2}\text{cis}\!\left(-\tfrac{\pi}{4}\right)\)
Convert to Cartesian
\(\dfrac{1}{\sqrt2}\!\left(\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}i\right)=\tfrac{1}{2}-\tfrac{1}{2}i\)
Answer
\(\dfrac{1}{2}-\dfrac{1}{2}i\)
Question 3
Using De Moivre's theorem, derive the identities for \(\cos 3\theta\) and \(\sin 3\theta\) in terms of \(\cos\theta\) and \(\sin\theta\).
Expand \((\cos\theta+i\sin\theta)^3\)
\(\cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\)
Simplify
\(=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)\)
Answer
\(\cos3\theta=4\cos^3\theta-3\cos\theta\); \(\sin3\theta=3\sin\theta-4\sin^3\theta\)
Question 4
Solve \(z^2-(3+i)z+(2+3i)=0\) by the quadratic formula, giving answers in Cartesian form.
Discriminant
\(\Delta=(3+i)^2-4(2+3i)=9+6i-1-8-12i=(-6i)\)
Square root of \(-6i\)
Let \(\sqrt{-6i}=a+bi\): \(a^2-b^2=0\) and \(2ab=-6\Rightarrow a=b,\;2a^2=6\Rightarrow a=\sqrt3\Rightarrow\sqrt{-6i}=\pm(\sqrt3-\sqrt3 i)\)
Solutions
\(z=\dfrac{(3+i)\pm(\sqrt3-\sqrt3 i)}{2}\)
Answer
\(z=\dfrac{3+\sqrt3}{2}+\dfrac{1-\sqrt3}{2}i\) or \(z=\dfrac{3-\sqrt3}{2}+\dfrac{1+\sqrt3}{2}i\)
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