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AI HL · AHL 6

Complex Numbers

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AHL 6.10

Adding sinusoidal functions with complex exponentials

AHL 6.10

Adding sinusoidal functions with complex exponentials

Target: I can write \(r\cos\theta\) as the real part and \(r\sin\theta\) as the imaginary part of \(re^{i\theta}\). (+4 more)

By the end of this section

Key points

A wave as a complex number

  • \(e^{i\theta} = \cos\theta + i\sin\theta\), so \(r\cos\theta = \mathrm{Re}(re^{i\theta})\) and \(r\sin\theta = \mathrm{Im}(re^{i\theta})\).
  • Use Re for cosines and Im for sines. Use the same one for every term in the sum.
  • \(e^{(\omega t + \phi)i} = e^{\omega ti} \times e^{\phi i}\). Waves of the same frequency have the same \(e^{\omega ti}\), so only the constants differ.

Adding two waves of the same frequency

  • Write each term as \(\mathrm{Re}\) (or \(\mathrm{Im}\)) of \(Ae^{(\omega t + \phi)i}\). Take out the common \(e^{\omega ti}\).
  • Add the constants \(A_1e^{\phi_1 i} + A_2e^{\phi_2 i}\) as complex numbers. Convert the answer to \(re^{\theta i}\) with the calculator, in radians.
  • Put the angle in \(-\pi < \theta \le \pi\): if the real part is negative, check the quadrant. The result is \(r\cos(\omega t + \theta)\) or \(r\sin(\omega t + \theta)\).
  • The frequency does not change. The amplitude \(r\) is not the sum of the amplitudes, unless the waves are in phase.
  • \(a\cos x + b\sin x = \mathrm{Re}\left((a - bi)e^{xi}\right)\). The constant is \(a - bi\), with a minus sign. \(B\) is the argument of \(a - bi\).

Reading the answer

  • The greatest value of \(A\cos(\omega t + B)\) is \(A\), and the least is \(-A\). The greatest voltage in a circuit is the amplitude.
  • \(A\cos(\omega t + B)\) is greatest when \(\omega t + B = 2k\pi\), least when \(\omega t + B = \pi + 2k\pi\) and zero when \(\omega t + B = \tfrac{\pi}{2} + k\pi\).
  • \(A\sin(\omega t + B)\) is greatest when \(\omega t + B = \tfrac{\pi}{2} + 2k\pi\) and zero when \(\omega t + B = k\pi\). Choose the smallest \(t \ge 0\) for the first time.
  • To find the share of one wave at a time \(t\): one wave's value divided by the total at that time, times 100. Put the numbers in, in radians.

On the calculator

Converting the sum of constants to polar form

  • Set the calculator to radians and to Complex mode.
  • Enter the sum of the constants, for example \(6e^{0.5i} + 4e^{2i}\), and read it as \(a + bi\).
  • Convert the answer to polar form \(re^{\theta i}\). Check that \(\theta\) is between \(-\pi\) and \(\pi\).

Screenshot and video to come

Solving H(t) = 0 over an interval

  • Enter the combined wave as a function of \(t\).
  • Solve \(H(t) = 0\) with the interval set to the times in the question, for example \(0 \le t \le 8\).
  • Read every solution in the interval. Zeros of a sinusoid are equally spaced, so check the gaps.

Screenshot and video to come

Checking a graph and its maximum

  • Graph the combined wave with the window set to the interval in the question.
  • Use the maximum, minimum and zero tools to check the coordinates you found by hand.
  • Compare the first maximum with \(t = \dfrac{-B}{\omega}\) for a cosine, adding a full period if that is negative.

Screenshot and video to come

Explore: adding two waves with complex numbers

Each wave A cos(t + φ) is the real part of A eφi eti, so it is the arrow A eφi on the Argand diagram: length A, angle φ. Drag the tips of the two arrows (or type values). Before you look at the orange wave, add the arrows in your head and say how big the sum will be.

The orange wave is the sum. Dragging a tip moves the angle in steps of π/12; type a phase for any other value. Numbers are rounded to 3 significant figures, but the sums are worked from the unrounded values.

  1. Give both arrows the same angle. What is the amplitude of the sum? Why does it equal A₁ + A₂ here and nowhere else?
  2. Turn z₂ until the orange wave is as small as you can make it. How far apart are φ₁ and φ₂ then? Compare the amplitude with the difference of A₁ and A₂.
  3. Make A₂ bigger and bigger. Which way does the arrow of the sum turn, and what happens to the phase of the sum wave?
  4. The dots on the left edge of the waves are the values at t = 0. Which part of each arrow on the Argand diagram do they match?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Adding two cosines

Write \(6\cos(3t + 0.5) + 4\cos(3t + 2)\) in the form \(A\cos(3t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

Working
  1. Real part of exponentials

    \(6\cos(3t + 0.5) + 4\cos(3t + 2) = \mathrm{Re}\left(6e^{(3t + 0.5)i} + 4e^{(3t + 2)i}\right)\)

    A cosine is the real part of a complex exponential. Write every term as \(\mathrm{Re}(Ae^{\theta i})\).

  2. Take out the common factor

    \(= \mathrm{Re}\left(e^{3ti}(6e^{0.5i} + 4e^{2i})\right)\)

    Every term has the same \(e^{3ti}\). Take it out, so the rest is constants.

  3. Add the constants

    From the calculator: \(6e^{0.5i} + 4e^{2i} = 3.60 + 6.51i = 7.44e^{1.07i}\)

    Add the two constants on the calculator, then change the answer to polar form. The radius is the new amplitude.

  4. Take the real part

    \(= \mathrm{Re}\left(7.44e^{(3t + 1.07)i}\right) = 7.44\cos(3t + 1.07)\)

    Combine the exponents and take the real part. The frequency 3 is unchanged.

Video walkthrough coming
Showing, then combining sines

(a) Show that \(4\sin(2t + 0.8) + 3\sin(2t + 3) = \mathrm{Im}\left(4e^{(2t + 0.8)i} + 3e^{(2t + 3)i}\right)\).<br>(b) Hence write \(4\sin(2t + 0.8) + 3\sin(2t + 3)\) in the form \(A\sin(2t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

Working
  1. Show: one term

    \(e^{\theta i} = \cos\theta + i\sin\theta \text{, so } \mathrm{Im}\left(4e^{(2t+0.8)i}\right) = 4\sin(2t + 0.8)\)

    The imaginary part of \(e^{\theta i}\) is \(\sin\theta\). The 4 is real, so it stays.

  2. Show: the sum

    \(\mathrm{Im}(z_1 + z_2) = \mathrm{Im}(z_1) + \mathrm{Im}(z_2) = 4\sin(2t + 0.8) + 3\sin(2t + 3)\)

    The imaginary part of a sum is the sum of the imaginary parts. This proves the statement.

  3. Factor

    \(= \mathrm{Im}\left(e^{2ti}(4e^{0.8i} + 3e^{3i})\right)\)

    Hence: take out \(e^{2ti}\) from both terms.

  4. Calculator

    From the calculator: \(4e^{0.8i} + 3e^{3i} = -0.183 + 3.29i = 3.30e^{1.63i}\)

    Convert the constant to polar form. Check the quadrant: \(\theta\) must be between \(-\pi\) and \(\pi\).

  5. Imaginary part

    \(= \mathrm{Im}\left(3.30e^{(2t + 1.63)i}\right) = 3.30\sin(2t + 1.63)\)

    Take the imaginary part, because the original terms are sines.

Video walkthrough coming
a cos x + b sin x

Write \(8\cos x + 15\sin x\) in the form \(A\cos(x + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

Working
  1. Choose a complex number

    \(\mathrm{Re}\left((8 - 15i)(\cos x + i\sin x)\right) = 8\cos x + 15\sin x\)

    Multiply out: the real part is \(8\cos x - (-15)\sin x\). So the sine coefficient must be minus the imaginary part.

  2. Write it as a real part

    \(8\cos x + 15\sin x = \mathrm{Re}\left((8 - 15i)e^{xi}\right)\)

    Now there is one constant to convert.

  3. Calculator

    From the calculator: \(8 - 15i = 17e^{-1.08i}\)

    The point is in the fourth quadrant, so the argument is negative.

  4. Take the real part

    \(= \mathrm{Re}\left(17e^{(x - 1.08)i}\right) = 17\cos(x - 1.08)\)

    A is 17 and B is negative.

Video walkthrough coming
Maximum, minimum and zero

The graph shows \(y = 5\cos x + 3\cos(x - 1.2)\) for \(0 \le x \le 6.5\). P is the maximum, Q is the minimum and R is the right-hand x-intercept.<br>(a) Write \(y\) in the form \(A\cos(x + B)\).<br>(b) Hence find the coordinates of P, Q and R.

123456−8−6−4−22468xyPQR
Working
  1. Combine

    From the calculator: \(5 + 3e^{-1.2i} = 6.09 - 2.80i = 6.70e^{-0.431i}\)

    Combine as before. This is a sum of two cosines with \(5e^{0i}\) and \(3e^{-1.2i}\).

  2. The form

    \(y = 6.70\cos(x - 0.431)\)

    The amplitude is A and the graph is shifted to the right by \(-B\).

  3. Maximum and minimum

    \(P: x - 0.431 = 0 \Rightarrow x = 0.431,\ y = 6.70\)<br>\(Q: x - 0.431 = \pi \Rightarrow x = 3.57,\ y = -6.70\)

    Cosine is largest when its angle is 0 and smallest when its angle is \(\pi\).

  4. 123456−8−6−4−22468xyPQR
    The x-intercept

    \(R: x - 0.431 = \tfrac{3\pi}{2} \Rightarrow x = 5.14,\ y = 0\)

    Cosine is zero when the angle is \(\tfrac{\pi}{2}\) or \(\tfrac{3\pi}{2}\). The right-hand intercept is the larger one.

Video walkthrough coming
Greatest voltage

Two alternating voltage sources are connected in one circuit. At time \(t\) milliseconds the voltages, in volts, are \(V_A = 12\cos(50t)\) and \(V_B = 9\cos(50t + 1)\). Find the greatest total voltage in the circuit.

Working
  1. Combine

    From the calculator: \(12 + 9e^{1i} = 16.9 + 7.57i = 18.5e^{0.422i}\)

    Both terms have the form \(A\cos(50t + \phi)\), so add \(12 + 9e^{i}\).

  2. Total voltage

    \(V = 18.5\cos(50t + 0.422)\)

    Take the real part of the exponential.

  3. Greatest value

    \(\text{greatest } V = 18.5 \text{ volts}\)

    Cosine has greatest value 1, so the greatest voltage is the amplitude A.

Video walkthrough coming
Zeros, first maximum and percentage

A buoy floats in a harbour. At time \(t\) seconds it is lifted by waves from a ferry, height \(0.4\sin(3t)\) m, and from a fishing boat, height \(0.25\sin(3t + 2)\) m.<br>(a) Write the total height \(H(t)\) in the form \(A\sin(3t + B)\).<br>(b) Find the times in the first 8 seconds when \(H(t) = 0\).<br>(c) Find the first time at which \(H\) is greatest.<br>(d) At that time, what percentage of the wave is due to the ferry?

Working
  1. Combine (a)

    From the calculator: \(0.4 + 0.25e^{2i} = 0.296 + 0.227i = 0.373e^{0.655i}\)

    Add the constants \(0.4 + 0.25e^{2i}\) and convert to polar form.

  2. The total wave (a)

    \(H(t) = 0.373\sin(3t + 0.655)\)

    Imaginary part, because the waves are sines.

  3. Zeros (b)

    Solve for \(t\) on \(0 \le t \le 8\): \(0.373\sin(3t + 0.655) = 0\) From the calculator: \(t = 0.829, 1.88, 2.92, 3.97, 5.02, 6.06, 7.11\)

    A sine is zero when its angle is a multiple of \(\pi\). The zeros are \(\pi / 3\) apart.

  4. First maximum (c)

    \(3t + 0.655 = \tfrac{\pi}{2} \Rightarrow t = 0.305\)

    A sine is greatest when its angle is \(\tfrac{\pi}{2}\). Choose the smallest positive \(t\).

  5. Percentage (d)

    \(\dfrac{0.4\sin(3 \times 0.305)}{H(0.305)} \times 100 = \dfrac{0.317}{0.373} \times 100 = 85.0\%\)

    Ferry wave divided by the total wave at that time, times 100.

Video walkthrough coming