Complex Numbers VII: Converting Between Forms
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
Convert the following to exponential (Euler) form \(re^{i heta}\):
(a)\(z=1+i\)
(b)\(z=-4\)
(c)\(z=2-2\sqrt{3}i\)
(a)
Answer
\(\sqrt{2}\,e^{i\pi/4}\)
(c)
\(r=4,\; heta=- frac{\pi}{3}\)
Question 2
Convert from exponential form to Cartesian:
(a)\(3e^{i\pi/2}\)
(b)\(5e^{-i\pi/6}\)
(c)\(2e^{i\pi}\)
(b)
Answer
\(5\!\left(\dfrac{\sqrt3}{2}-\dfrac{1}{2}i
ight)=\dfrac{5\sqrt3}{2}-\dfrac{5}{2}i\)
Question 3
Using Euler's formula \(e^{i heta}=\cos heta+i\sin heta\), derive the expressions for \(\cos heta\) and \(\sin heta\) in terms of exponentials.
Add and subtract
\(e^{i heta}+e^{-i heta}=2\cos heta\Rightarrow\cos heta=\dfrac{e^{i heta}+e^{-i heta}}{2}\)
Similarly
\(\sin heta=\dfrac{e^{i heta}-e^{-i heta}}{2i}\)
Answer
\(\cos heta=\dfrac{e^{i heta}+e^{-i heta}}{2};\quad\sin heta=\dfrac{e^{i heta}-e^{-i heta}}{2i}\)
Question 4
Verify Euler's identity \(e^{i\pi}+1=0\) by substituting \( heta=\pi\) into the formula. Hence explain why this is considered remarkable.
Substitute
\(e^{i\pi}=\cos\pi+i\sin\pi=-1+0=-1\Rightarrow e^{i\pi}+1=0\) ✓
Explanation
Euler's identity connects five of the most fundamental constants in mathematics: \(e,\;i,\;\pi,\;1,\;0\) — each from a completely different area — in a single elegant equation.
Question 5
Simplify \(\dfrac{e^{i\pi/3}\cdot e^{i\pi/6}}{e^{i\pi/4}}\), giving your answer in exact Cartesian form.
Combine exponents
\(e^{i(\pi/3+\pi/6-\pi/4)}=e^{i(4\pi/12+2\pi/12-3\pi/12)}=e^{i\cdot3\pi/12}=e^{i\pi/4}\)
Answer
\(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i\)
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