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AI HL · AHL 6

Complex Numbers

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AHL 6.11

Mixed practice: polar and exponential form

AHL 6.11

Mixed practice: polar and exponential form

Target: I can convert a complex number between Cartesian, modulus-argument and exponential form, in any quadrant, with the argument in the interval asked for. (+5 more)

By the end of this section

Key points

Changing between forms

  • Sketch the point first. \(|z| = \sqrt{x^2 + y^2}\), and the argument comes from the quadrant and the angle with the real axis: first \(\theta\), second \(\pi - \theta\), third \(\pi + \theta\), fourth \(2\pi - \theta\).
  • Read the interval the question gives. \(0 \le \theta < 2\pi\) and \(-\pi < \theta \le \pi\) give different answers in the third and fourth quadrants.
  • \(r\,\mathrm{cis}\,\theta = re^{i\theta} = r\cos\theta + ir\sin\theta\), so \(x = r\cos\theta\) and \(y = r\sin\theta\).
  • Work without a calculator when the angle is a standard one, then check with the calculator.

Products, quotients and powers

  • Multiply: multiply the moduli and add the arguments. Divide: divide the moduli and subtract the arguments.
  • Power: \((r\,\mathrm{cis}\,\theta)^n = r^n\,\mathrm{cis}\,n\theta\) and \((re^{i\theta})^n = r^ne^{in\theta}\). Raise the modulus to the power; multiply the argument.
  • After combining, bring the argument back into the interval asked for by adding or subtracting \(2\pi\).
  • To give an answer in Cartesian form, work out \(r\cos\theta\) and \(r\sin\theta\) at the end. Do not round before then.

Using the rules

  • \(\arg(zw) = \arg z + \arg w\) and \(|zw| = |z||w|\), up to a multiple of \(2\pi\) in the argument.
  • \(z^n\) is real when \(n\arg z\) is a multiple of \(\pi\). \(z^n = i\) when \(n\arg z = \frac{\pi}{2} + 2k\pi\). Try \(k = 0, 1, 2, \ldots\) until \(n\) is a whole number.
  • \(z^2\) has modulus \(|z|^2\) and twice the argument. \(iz\) is \(z\) rotated \(\frac{\pi}{2}\) anticlockwise about the origin.
  • From \(zw\) and \(\frac{z}{w}\): multiply the moduli to get \(|z|^2\), and add the arguments to get \(2\arg z\).

Sums and sinusoids

  • Complex numbers in polar form cannot be added directly. Add in Cartesian form, then convert the sum to polar form.
  • \(A\sin\theta + B\sin(\theta + \phi) = \mathrm{Im}\left(e^{\theta i}(A + Be^{\phi i})\right)\). The modulus and argument of \(A + Be^{\phi i}\) give the amplitude and phase of the sum.
  • A population model \(C_1 + a_1\sin(bt) + C_2 + a_2\sin(bt + \phi)\) has the constants added separately: \(C = C_1 + C_2\), and the sinusoids combined.

On the calculator

Converting between Cartesian and polar form

  • Set the calculator to radians and to Complex mode.
  • Enter the number as \(x + iy\) and convert it to polar form to read \(r\) and \(\theta\). Check which interval the calculator gives, and add \(2\pi\) if the question needs \(0 \le \theta < 2\pi\).
  • To go the other way, work out \(r\cos\theta\) for the real part and \(r\sin\theta\) for the imaginary part.

Screenshot and video to come

Checking a product or power in Cartesian form

  • Do the question by hand first, without the calculator.
  • Combine the modulus and the argument by hand, then enter \(r\cos\theta\) and \(r\sin\theta\) for the real and imaginary parts.
  • Enter the whole expression in Complex mode and compare the answer with your working.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Third quadrant to modulus-argument form

Write \(z = -5 - 5\sqrt{3}\,i\) in the form \(r\,\mathrm{cis}\,\theta\) with \(0 \le \theta < 2\pi\). Work without a calculator first, then check on the calculator.

−12−10−8−6−4−224−12−10−8−6−4−224ReImz
Working
  1. Modulus

    \(|z| = \sqrt{(-5)^2 + (- 5 \sqrt{3})^2} = 10\)

    Pythagoras on the real and imaginary parts.

  2. −12−10−8−6−4−224−12−10−8−6−4−224ReImz
    Sketch and angle

    The point is in the third quadrant. The angle with the real axis is \(\frac{\pi}{3}\), so \(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\)

    Sketch the point first. Both parts are negative, so the point is in the third quadrant.

  3. Write the form

    \(z = 10\,\mathrm{cis}\left(\frac{4\pi}{3}\right)\)

    Check on the calculator: \(10\cos\frac{4\pi}{3} = -5\) and \(10\sin\frac{4\pi}{3} = -5\sqrt{3}\).

Video walkthrough coming
Products and quotients in cis form

Write each in the form \(r\,\mathrm{cis}\,\theta\) with \(0 < \theta \le 2\pi\).<br>(a) \((4\,\mathrm{cis}\left(\frac{5\pi}{4}\right))(3\,\mathrm{cis}\left(\frac{7\pi}{6}\right))\)<br>(b) \(\dfrac{12\,\mathrm{cis}\left(\frac{\pi}{6}\right)}{3\,\mathrm{cis}\left(\frac{2\pi}{3}\right)}\)

Working
  1. (a) Multiply the moduli

    \(|z| = 4 \times 3 = 12\)

    To multiply, multiply the moduli.

  2. (a) Add the arguments

    \(\arg = \frac{5\pi}{4} + \frac{7\pi}{6} = \frac{29\pi}{12}\) This is outside the interval, so subtract \(2\pi\): \(\frac{5\pi}{12}\)

    To multiply, add the arguments. The interval is \(0 < \theta \le 2\pi\), so reduce if the sum is too large.

  3. (b) Divide

    \(|z| = 12 \div 3 = 4\)<br>\(\arg = \frac{\pi}{6} - \frac{2\pi}{3} = -\frac{\pi}{2}\) This is outside the interval, so add \(2\pi\): \(\frac{3\pi}{2}\)

    To divide, divide the moduli and subtract the arguments. A negative angle needs \(2\pi\) added.

  4. (b) Result

    \(= 4\,\mathrm{cis}\left(\frac{3\pi}{2}\right)\)

    The argument is now in the interval \(0 < \theta \le 2\pi\).

Video walkthrough coming
Powers and products in Cartesian form

Write \(\left(3\,\mathrm{cis}\left(\frac{\pi}{4}\right)\right)^{3}\left(2\,\mathrm{cis}\left(\frac{\pi}{5}\right)\right)^{2}\) in Cartesian form, to 3 significant figures.

Working
  1. The modulus

    \(|z| = 3^{3} \times 2^{2} = 108\)

    Raise each modulus to its power, then multiply.

  2. The argument

    \(\arg = 3 \times \frac{\pi}{4} + 2 \times \frac{\pi}{5} = \frac{23\pi}{20}\)

    Multiply each argument by its power, then add.

  3. Polar form

    \(= 108\,\mathrm{cis}\left(\frac{23\pi}{20}\right)\)

    Keep the angle in \([0, 2\pi)\).

  4. Cartesian form

    From the calculator: \(108\cos(\frac{23\pi}{20}) = -96.2,\quad 108\sin(\frac{23\pi}{20}) = -49.0\)<br>\(z = -96.2 - 49.0i\)

    Real part \(r\cos\theta\), imaginary part \(r\sin\theta\).

Video walkthrough coming
Products and quotients in exponential form

Write each in the form \(re^{i\theta}\) with \(-\pi < \theta \le \pi\).<br>(a) \(5e^{\frac{2\pi i}{3}} \times 4e^{\frac{3\pi i}{4}}\)<br>(b) \(18e^{\frac{\pi i}{4}} \div 6e^{-\frac{\pi i}{3}}\)

Working
  1. (a) Moduli and angles

    \(|z| = 5 \times 4 = 20\)<br>\(\text{angle} = \frac{2\pi}{3} + \frac{3\pi}{4} = \frac{17\pi}{12}\) Subtract \(2\pi\) to bring it into \(-\pi < \theta \le \pi\): \(-\frac{7\pi}{12}\)

    In exponential form add the exponents: \(e^{ia} \times e^{ib} = e^{i(a + b)}\).

  2. (a) Result

    \(= 20e^{-\frac{7\pi i}{12}}\)

    The angle is now in \(-\pi < \theta \le \pi\).

  3. (b) Divide

    \(|z| = 18 \div 6 = 3\)<br>\(\text{angle} = \frac{\pi}{4} - \left(-\frac{\pi}{3}\right) = \frac{7\pi}{12}\)

    To divide, divide the moduli and subtract the exponents.

  4. (b) Result

    \(= 3e^{\frac{7\pi i}{12}}\)

    The angle is already in the interval.

Video walkthrough coming
A power in exponential form

Write \(\left(2e^{\frac{\pi i}{6}}\right)^5\) in the form \(x + iy\), without a calculator.

Working
  1. Raise to the power

    \(\left(2e^{\frac{\pi i}{6}}\right)^5 = 2^5e^{\frac{5\pi i}{6}} = 32e^{\frac{5\pi i}{6}}\)

    The modulus is raised to the power and the angle is multiplied by it.

  2. Cartesian form

    \(= 32\left(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\right) = - 16 \sqrt{3} + 16 i\)

    Use the exact values \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\) and \(\sin\frac{5\pi}{6} = \frac{1}{2}\).

Video walkthrough coming
The smallest power that equals i

Let \(z = \mathrm{cis}\left(\frac{3\pi}{14}\right)\). Find the smallest positive integer \(n\) for which \(z^n = i\).

Working
  1. The argument of z^n

    \(z^n = \mathrm{cis}\left(\tfrac{3n\pi}{14}\right)\)

    The modulus is 1, so only the argument matters.

  2. What i needs

    \(\tfrac{3n\pi}{14} = \tfrac{\pi}{2} + 2k\pi \Rightarrow 3n = 7 + 28k\)

    \(i\) has argument \(\frac{\pi}{2}\), and any multiple of \(2\pi\) can be added.

  3. Try k

    \(k = 0: n = \frac{7}{3}; \ k = 1: n = \frac{35}{3}; \ k = 2: n = 21\)

    Stop at the first whole number.

  4. Answer

    \(n = 21\)

    Check: \(21 \times \frac{3\pi}{14} = \frac{9\pi}{2} = 4\pi + \frac{\pi}{2}\).

Video walkthrough coming
A sum with an exponential term

(a) Find the modulus and argument of \(7 + 3e^{2i}\).<br>(b) Hence write \(7\sin\theta + 3\sin(\theta + 2)\) in the form \(A\sin(\theta + \alpha)\), where \(A > 0\) and \(-\pi < \alpha \le \pi\).

Working
  1. Cartesian form

    From the calculator: \(7 + 3e^{2i} = 5.75 + 2.73i\)

    Add the real parts and the imaginary parts. A sum cannot be added in polar form.

  2. Modulus and argument

    \(|z| = 6.37,\ \arg z = 0.443\)

    Convert the sum to polar form.

  3. The sinusoid

    \(7\sin\theta + 3\sin(\theta + 2) = \mathrm{Im}\left(e^{\theta i}(7 + 3e^{2i})\right) = 6.37\sin(\theta + 0.443)\)

    Take out the common factor \(e^{\theta i}\), and take the imaginary part. This is the method of the last lesson.

Video walkthrough coming
Sketching two voltages and their total

Two AC voltage sources are connected in a circuit. The voltages, in volts, at time \(t\) seconds are \(V_1 = 6\cos(20t)\) and \(V_2 = 9\cos(20t + 4)\).<br>(a) Sketch \(V_1\) and \(V_2\) for \(0 \le t \le 0.5\).<br>(b) Find the total voltage \(V = V_1 + V_2\) in the form \(A\cos(20t + c)\).

0.10.20.30.40.5−10−8−6−4−2246810tVV1V2V
Working
  1. 0.10.20.30.40.5−12−8−44812tV
    (a) The two graphs

    \(V_1 = 6\cos(20t)\): amplitude 6, period \(\tfrac{2\pi}{20} = 0.314\), starts at 6.<br>\(V_2 = 9\cos(20t + 4)\): amplitude 9, same period, starts at -5.88.

    Same frequency, so the same period. The 4 in \(V_2\) shifts its graph to the left by \(\frac{4}{20} = 0.2\). In the graph, teal is \(V_1\), orange is \(V_2\) and blue is the total from part (b).

  2. (b) The sum

    \(= \mathrm{Re}\left(e^{20ti}(6 + 9e^{4i})\right)\)<br>From the calculator: \(6 + 9e^{4i} = 0.117 - 6.81i = 6.81e^{-1.55i}\)

    Take out \(e^{20ti}\) and add the constants.

  3. (b) The total voltage

    \(= \mathrm{Re}\left(6.81e^{(20t - 1.55)i}\right) = 6.81\cos(20t - 1.55)\)

    Check against the sketch: the total has amplitude \(A\), between 3 and 15.

Video walkthrough coming