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9709 · 3.9

Complex numbers — practice questions

Practice and worked examples for 9709 Complex numbers. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Let the complex number u=3+iu = -\sqrt{3} + i. <br> (i) Find the modulus and argument of uu. <br> (ii) Express uu in exponential form.

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Given u=3+iu = -\sqrt{3} + i. This is in the form x+iyx+iy with x=3x = -\sqrt{3} and y=1y = 1. <br><br> (i) Modulus and Argument <br> First, calculate the modulus: <br> u=(3)2+12=3+1=4=2|u| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2. [M1, A1] <br><br> Next, find the argument θ\theta. Let's first find the related acute angle α\alpha: <br> α=tan1yx=tan113=tan1(13)=π6\alpha = \tan^{-1} \left| \frac{y}{x} \right| = \tan^{-1} \left| \frac{1}{-\sqrt{3}} \right| = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}. [M1] <br><br> Now, sketch an Argand diagram. The point (3,1)(-\sqrt{3}, 1) is in the second quadrant. <br> For the second quadrant, the principal argument is θ=πα\theta = \pi - \alpha. <br> arg(u)=θ=ππ6=5π6\arg(u) = \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}. [A1] <br><br> (ii) Exponential Form <br> The exponential form is z=reiθz = re^{i\theta}. We have r=2r=2 and θ=5π6\theta = \frac{5\pi}{6}. <br> So, u=2ei(5π/6)u = 2e^{i(5\pi/6)}. [B1 ft]

Worked example 2

Use De Moivre's theorem to find (1i)8(1 - i)^8.

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To use De Moivre's theorem, we must first convert z=1iz = 1 - i into polar or exponential form. <br><br> Step 1: Find modulus and argument of z=1iz=1-i. <br> x=1,y=1x=1, y=-1. <br> Modulus: z=12+(1)2=2|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}. [M1] <br> Argument: The point (1,1)(1, -1) is in the fourth quadrant. <br> θ=tan1(11)=π4\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}. [M1] <br> So, z=2(cos(π4)+isin(π4))z = \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right). <br><br> Step 2: Apply De Moivre's Theorem. <br> We want to find z8z^8. <br> z8=[2(cos(π4)+isin(π4))]8z^8 = \left[ \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right) \right]^8 <br> z8=(2)8(cos(8×π4)+isin(8×π4))z^8 = (\sqrt{2})^8 \left( \cos(8 \times -\frac{\pi}{4}) + i\sin(8 \times -\frac{\pi}{4}) \right) [M1 for applying theorem] <br> (2)8=(21/2)8=24=16(\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16. <br> 8×π4=2π8 \times -\frac{\pi}{4} = -2\pi. <br> So, z8=16(cos(2π)+isin(2π))z^8 = 16 (\cos(-2\pi) + i\sin(-2\pi)). <br><br> Step 3: Convert back to Cartesian form. <br> We evaluate the trigonometric functions: <br> cos(2π)=1\cos(-2\pi) = 1 <br> sin(2π)=0\sin(-2\pi) = 0 <br> Therefore, z8=16(1+i0)=16z^8 = 16(1 + i \cdot 0) = 16. [A1] <br> The final answer is 16.