Complex Numbers II: Modulus & Argument
IB Mathematics AA · HL · Shadow Worksheet
Practice
Question 1
Find the modulus and argument (in radians) of the following:
(a)\(z=3+3i\)
(b)\(z=-\sqrt{3}+i\)
(c)\(z=-2-2i\)
(a)
Modulus and argument
\(|z|=3\sqrt{2};\;\arg(z)=\arctan(1)=\tfrac{\pi}{4}\)
Answer
\(|z|=3\sqrt{2},\;\arg z=\dfrac{\pi}{4}\)
(b)
Q2: \(|z|=2\)
\(\arg z=\pi-\arctan\tfrac{1}{\sqrt3}=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}\)
Answer
\(|z|=2,\;\arg z=\dfrac{5\pi}{6}\)
(c)
Q3: \(|z|=2\sqrt{2}\)
\(\arg z=-\pi+\tfrac{\pi}{4}=-\tfrac{3\pi}{4}\)
Answer
\(|z|=2\sqrt{2},\;\arg z=-\dfrac{3\pi}{4}\)
Question 2
Write the following in polar form \(r(\cos\theta+i\sin\theta)\):
(a)\(z=1+i\)
(b)\(z=-\sqrt{3}-i\)
(a)
Answer
\(\sqrt{2}\!\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)\)
(b)
Q3: \(r=2,\;\theta=-\tfrac{5\pi}{6}\)
Answer
\(2\!\left(\cos\!\left(-\dfrac{5\pi}{6}\right)+i\sin\!\left(-\dfrac{5\pi}{6}\right)\right)\)
Question 3
Convert from polar to Cartesian form:
(a)\(4\!\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)\)
(b)\(3\,\text{cis}\!\left(-\dfrac{\pi}{2}\right)\)
(a)
Answer
\(4\!\left(\tfrac{1}{2}+\tfrac{\sqrt3}{2}i\right)=2+2\sqrt{3}i\)
Question 4
Given \(z_1=2\,\text{cis}\!\left(\dfrac{\pi}{6}\right)\) and \(z_2=3\,\text{cis}\!\left(\dfrac{\pi}{4}\right)\), find \(z_1z_2\) and \(\dfrac{z_1}{z_2}\) in polar form.
Multiplication and division rules
Multiply moduli, add/subtract arguments
Answer
\(z_1z_2=6\,\text{cis}\!\left(\dfrac{5\pi}{12}\right)\); \(\dfrac{z_1}{z_2}=\dfrac{2}{3}\,\text{cis}\!\left(-\dfrac{\pi}{12}\right)\)
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