AI HL · AHL 6
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Powers and de Moivre's formula
AHL 6.9
Screenshot and video to come
Drag z (orange) or use the sliders for its modulus and argument. The teal points are z, z², z³ and so on, up to the power you choose. Predict where the next point goes before you move the slider for n.
The argument goes up in steps of π/24. Select z (Tab): the left and right arrow keys change its argument, the up and down keys change its modulus. The dashed circle has radius 1. The black arrow goes to the power you chose; the gold arc is its argument.
\(z = 2e^{i\frac{\pi}{18}}\). Write \(z^6\) in the form \(x + iy\). Give the real part \(x\).
Have a go first. Then open the working one line at a time.
\(|z^6| = 2^6 = 64\)
The power applies to the modulus.
\(\arg(z^6) = 6 \times \frac{\pi}{18} = \frac{\pi}{3}\)
Same rule as \((e^{i\theta})^n = e^{in\theta}\).
\(z^6 = 64e^{i\frac{\pi}{3}} = 64\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 32+32 \sqrt{3}i\)
From the calculator, or the unit circle: \(\cos\frac{\pi}{3} = \frac{1}{2}\). So \(x = 32\).
\(z = \sqrt{3} - i\). (a) Find the modulus and argument of \(z\), with \(0 \le \arg z < 2\pi\). (b) Hence write \(z^5\) in the form \(x + iy\). Give the imaginary part \(y\).
Have a go first. Then open the working one line at a time.
\(|z| = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2\)
Sketch the point first: it is below the real axis.
\(\text{The point is in the fourth quadrant. The acute angle is } \arctan\frac{1}{\sqrt{3}} = \frac{\pi}{6}, \text{ so } \arg z = \frac{11\pi}{6}\)
Always check the quadrant before you choose the angle.
\(z = 2e^{i\frac{11\pi}{6}}\)
Now the power is a single step.
\(z^5 = \left(2e^{i\frac{11\pi}{6}}\right)^5 = 2^{5}e^{i \times 5 \times \frac{11\pi}{6}} = 32e^{i\frac{55\pi}{6}}\)
Modulus to the power 5. Argument times 5.
\(\frac{55\pi}{6} - 8\pi = \frac{7\pi}{6}\) \(z^5 = 32e^{i\frac{7\pi}{6}} = 32\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right) = -16 \sqrt{3}-16i\)
The imaginary part is \(y = 32 \times \left(-\frac{1}{2}\right) = -16\).
\(w = 2\,\text{cis}\left(\frac{\pi}{8}\right)\). Use de Moivre's formula to write \(w^4\) in the form \(x + iy\). Give the imaginary part \(y\).
Have a go first. Then open the working one line at a time.
\((r\,\text{cis}\,\theta)^n = r^n\,\text{cis}(n\theta)\)
This is the exponential rule written in cis form.
\(w^4 = \left(2\,\text{cis}\left(\frac{\pi}{8}\right)\right)^4 = 2^{4}\,\text{cis}\left(4 \times \frac{\pi}{8}\right) = 16\,\text{cis}\left(\frac{\pi}{2}\right)\)
Modulus \(2^4 = 16\). Argument \(4 \times \frac{\pi}{8}\).
\(w^4 = 16\,\text{cis}\left(\frac{\pi}{2}\right) = 16\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 16i\)
\(\cos\frac{\pi}{2} = 0\) and \(\sin\frac{\pi}{2} = 1\), so \(y = 16\).
A geometric sequence has first term \(u_1 = 16\) and common ratio \(r = \frac{1}{2}(1 + i)\). (a) Find \(u_2\) and \(u_3\) in the form \(x + iy\). (b) Show that \(|u_1|, |u_2|, |u_3|, \dots\) is also a geometric sequence, and find its common ratio. (c) Find the sum of the infinite sequence \(|u_1|, |u_2|, |u_3|, \dots\).
Have a go first. Then open the working one line at a time.
\(u_2 = 16 \times \frac{1}{2}(1 + i) = 8 + 8i\)<br>\(u_3 = (8 + 8i) \times \frac{1}{2}(1 + i) = 4 + 4i + 4i - 4 = 8i\)
Multiply by \(r\) each time. The next two terms are \(-4 + 4i\) and \(-4\).
\(|u_1| = 16,\quad |u_2| = 8\sqrt{2},\quad |u_3| = 8\)
\(\dfrac{|u_2|}{|u_1|} = \dfrac{\sqrt{2}}{2}\) and \(\dfrac{|u_3|}{|u_2|} = \dfrac{\sqrt{2}}{2}\).
\(|u_{n+1}| = |u_n r| = |u_n||r|\), with \(|r| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{\sqrt{2}}{2}\)
The modulus of a product is the product of the moduli, so the ratio is the constant \(|r| = 0.707\).
\(|r| < 1\), so \(S_\infty = \dfrac{16}{1 - \frac{\sqrt{2}}{2}} = 54.6\)
The first term of the moduli is \(|u_1| = 16\). The points spiral in towards the origin.