Complex Numbers VI: Mixed
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
If \(z=2(\cos heta+i\sin heta)\), find \(z+ar{z}\) and \(zar{z}\).
Conjugate
\(ar{z}=2(\cos heta-i\sin heta)\)
Answer
\(z+ar{z}=4\cos heta\); \(zar{z}=|z|^2=4\)
Question 2
Show that if \(z=r\, ext{cis}\, heta\) then \(\dfrac{1}{z}=\dfrac{1}{r} ext{cis}(- heta)\). Hence find \(\dfrac{1}{1+i}\) without rationalising.
Polar of \(1+i\)
\(r=\sqrt2,\; heta= frac{\pi}{4}\Rightarrow\dfrac{1}{1+i}=\dfrac{1}{\sqrt2} ext{cis}\!\left(- frac{\pi}{4}
ight)\)
Convert to Cartesian
\(\dfrac{1}{\sqrt2}\!\left( frac{\sqrt2}{2}- frac{\sqrt2}{2}i
ight)= frac{1}{2}- frac{1}{2}i\)
Answer
\(\dfrac{1}{2}-\dfrac{1}{2}i\)
Question 3
Using De Moivre's theorem, derive the identities for \(\cos 3 heta\) and \(\sin 3 heta\) in terms of \(\cos heta\) and \(\sin heta\).
Expand \((\cos heta+i\sin heta)^3\)
\(\cos^3 heta+3\cos^2 heta(i\sin heta)+3\cos heta(i\sin heta)^2+(i\sin heta)^3\)
Simplify
\(=(\cos^3 heta-3\cos heta\sin^2 heta)+i(3\cos^2 heta\sin heta-\sin^3 heta)\)
Answer
\(\cos3 heta=4\cos^3 heta-3\cos heta\); \(\sin3 heta=3\sin heta-4\sin^3 heta\)
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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