Polar and exponential form

Express complex numbers in modulus–argument form r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta) and Euler's form reiθre^{i\theta}; multiply and divide using polar form; apply De Moivre's theorem to compute powers.

Worked examples

Converting to polar form

Straightforward

Problem

Write z=−1+3 iz = -1+\sqrt{3}\,i in polar form.

Multiplying and dividing in polar form

Moderate

Problem

Given z1=6eiπ/4z_1 = 6e^{i\pi/4} and z2=2eiπ/12z_2 = 2e^{i\pi/12}, find z1z2\dfrac{z_1}{z_2} in polar form and hence in Cartesian form.

Using De Moivre's theorem to evaluate a power

Challenging

Problem

Use De Moivre's theorem, (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta+i\sin\theta)^n = \cos(n\theta)+i\sin(n\theta), to compute (1+i)10(1+i)^{10}.

Practise

Q1·Straightforward
Find the modulus of z=3+iz = \sqrt{3}+i.
Q2·Straightforward
z1=4(cos⁡20°+isin⁡20°)z_1 = 4(\cos 20° + i\sin 20°) and z2=3(cos⁡70°+isin⁡70°)z_2 = 3(\cos 70° + i\sin 70°). Find ∣z1z2∣|z_1 z_2|.
Q3·Straightforward
Find the real part of z=2(cos⁡90°+isin⁡90°)z = 2(\cos 90°+i\sin 90°).
Q4·Moderate
Find the argument of z=3+iz = \sqrt{3}+i in degrees.
Q5·Moderate
z1=3(cos⁡40°+isin⁡40°)z_1 = 3(\cos 40°+i\sin 40°) and z2=2(cos⁡80°+isin⁡80°)z_2 = 2(\cos 80°+i\sin 80°). Find arg⁡(z1z2)\arg(z_1 z_2) in degrees.
Q6·Moderate
Find the real part of (1+i)8(1+i)^8 using De Moivre's theorem.
Q7·Moderate
Find the argument of z=−3+iz = -\sqrt{3}+i in degrees.
Q8·Moderate
z1=10(cos⁡60°+isin⁡60°)z_1 = 10(\cos 60°+i\sin 60°) and z2=2(cos⁡10°+isin⁡10°)z_2 = 2(\cos 10°+i\sin 10°). Find ∣z1z2∣\left|\dfrac{z_1}{z_2}\right|.
Q9·Challenging
Find the real part of (cos⁡18°+isin⁡18°)10(\cos 18°+i\sin 18°)^{10} using De Moivre's theorem.
Q10·Challenging
Find the real part of (1+i)12(1+i)^{12} using De Moivre's theorem.
Q11·Challenging
Find the imaginary part of (1−i)10(1-i)^{10} using De Moivre's theorem.
Q12·Challenging
z=2eiπ/6z = 2e^{i\pi/6} and w=3eiπ/3w = 3e^{i\pi/3}. Find Im(zw)\text{Im}(zw).
Q13·Challenging
Use De Moivre's theorem to expand (cos⁡θ+isin⁡θ)3(\cos\theta+i\sin\theta)^3 and hence prove that:
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta

✎ Work this one through on paper — proofs are self-assessed.