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AI HL · AHL 6

Complex Numbers

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AHL 6.8

Exponential form

AHL 6.8

Exponential form

Target: I can write a complex number in exponential form \(re^{i\theta}\) and say what \(r\) and \(\theta\) are. (+5 more)

By the end of this section

Key points

Exponential (Euler) form

  • A complex number can be written \(z = re^{i\theta}\), where \(r = |z|\) and \(\theta = \arg z\). Here \(e^{i\theta} = \cos\theta + i\sin\theta\), so \(re^{i\theta} = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta\). The angle is in radians.
  • Why it works: the rules for arguments are the laws of indices. Arguments add when you multiply, as in \(a^m \times a^n = a^{m+n}\). So the argument can be written as an exponent.
  • A number has many exponential forms, because \(\theta\) and \(\theta + 2\pi\) give the same point. Use \(0 \le \theta < 2\pi\) unless the question gives another range.
  • Special value: \(e^{i\pi} = -1\), so \(e^{i\pi} + 1 = 0\) (Euler's identity).

Multiplying and dividing

  • To multiply, multiply the numbers in front and add the powers: \(re^{i\alpha} \times se^{i\beta} = rs\,e^{i(\alpha + \beta)}\).
  • To divide, divide the numbers in front and subtract the powers: \(\dfrac{re^{i\alpha}}{se^{i\beta}} = \dfrac{r}{s}\,e^{i(\alpha - \beta)}\).
  • After adding or subtracting, bring the argument back into range by adding or subtracting \(2\pi\).
  • The conjugate of \(re^{i\theta}\) is \(re^{-i\theta}\).
  • The conjugate has the same modulus and the opposite argument, so \(zz^* = re^{i\theta} \times re^{-i\theta} = r^2\). The product is real because the arguments cancel.

Choosing the form

  • Use exponential form for products, quotients and powers. Use Cartesian form for sums and differences.
  • To change a Cartesian number into exponential form, find \(r\) and \(\theta\) as in the last lesson: sketch the point first so that the quadrant is right.
  • To change back, write \(re^{i\theta} = r\cos\theta + ir\sin\theta\).
  • A formula such as \(z_\theta = \frac{1}{3}\theta e^{i\theta}\) has a modulus that grows with the argument. Its points turn anticlockwise and move away from the origin: a spiral. Find \(|z_\theta|\) and \(\arg z_\theta\) from the formula, then plot.

On the calculator

Using exponential form on the calculator

  • In Complex mode with the angle set to radians, a number can be typed as \(re^{i\theta}\). Products and quotients are returned in the form \(x + iy\).
  • Use the calculator's tools for the polar angle and the magnitude to read off \(\theta\) and \(r\) from a Cartesian answer.
  • Your working is what is marked. Use the calculator to check it.

Screenshot and video to come

Explore: z on the Argand diagram

Drag z (orange) to any point. Read the modulus and the argument, and the three ways of writing z. Predict which quadrant, and what θ will be, before you read the working.

The point snaps to whole and half numbers. Select it (Tab) and use the arrow keys to move it.

  1. Drag z into the third quadrant. tan⁻¹ of y/x gives a small positive angle. Why is the argument not equal to that angle?
  2. Show z* and −z. What happens to the modulus? What happens to the argument of z*, and of −z?
  3. Put z on the negative real axis and switch between the two ranges. Which values of θ do you get? Why does the second range give π and not −π?
  4. Which points have an argument that is a multiple of π/4? What do their coordinates have in common?

Explore: multiplying turns and scales

Drag z (black) and w (teal). The orange arrow is zw. Predict its distance from O and its angle before you read the working. Or press a button to see what multiplying does to a whole shape.

The points snap to whole and half numbers. Select one (Tab) and use the arrow keys to move it. Angles are in radians, to 3 significant figures; the sums are worked from the unrounded values. The dashed circle has radius 1.

  1. Press "Multiply by i". Where do A, B and C go? What happens to the size of the triangle, and to its angles?
  2. Drag w onto the dashed circle. What does multiplying by w do to the distance of every point from O? What does it still do to the angle?
  3. Press "Multiply by i" twice in your head: i × i. Which button gives the same result? Check with the arguments: π/2 + π/2.
  4. Show z ÷ w. Predict its distance and angle first. Why is dividing by w the same as turning back and shrinking?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A product in exponential form

\(z = 9e^{i\frac{\pi}{4}}\) and \(w = 2e^{i\frac{2\pi}{3}}\). Write \(zw\) in exponential form. The argument of \(zw\) is \(\frac{a\pi}{12}\).

Working
  1. Add the powers

    \(zw = 9e^{i\frac{\pi}{4}} \times 2e^{i\frac{2\pi}{3}} = 18e^{i\left(\frac{\pi}{4} + \frac{2\pi}{3}\right)} = 18e^{i\frac{11\pi}{12}}\)

    Same rule as \(a^m \times a^n = a^{m+n}\). The numbers in front multiply.

  2. Check the range

    \(\frac{11\pi}{12}\) is between 0 and \(2\pi\)

    No change is needed.

  3. Write the answer

    \(zw = 18e^{i\frac{11\pi}{12}}\)

    Modulus \(9 \times 2 = 18\).

Video walkthrough coming
A quotient with a negative argument

\(z = 15e^{i\frac{\pi}{6}}\) and \(w = 3e^{i\frac{5\pi}{6}}\). Write \(\dfrac{z}{w}\) in exponential form with \(0 \le \theta < 2\pi\). The argument is \(\frac{k\pi}{3}\).

Working
  1. Subtract the powers

    \(\dfrac{z}{w} = \dfrac{15e^{i\frac{\pi}{6}}}{3e^{i\frac{5\pi}{6}}} = 5e^{i\left(\frac{\pi}{6} - \frac{5\pi}{6}\right)} = 5e^{i\left(-\frac{2\pi}{3}\right)}\)

    Same rule as \(\dfrac{a^m}{a^n} = a^{m-n}\). Divide the numbers in front.

  2. Bring it into range

    \(-\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}\)

    A negative argument is the same direction as the one \(2\pi\) larger.

  3. Write the answer

    \(\dfrac{z}{w} = 5e^{i\frac{4\pi}{3}}\)

    Modulus \(15 \div 3 = 5\).

Video walkthrough coming
A product on the imaginary axis

\(z = 3e^{i\frac{\pi}{5}}\) and \(w = 4e^{i\frac{3\pi}{10}}\). Write \(zw\) in exponential form, then in the form \(x + iy\). Give \(y\).

Working
  1. Add the powers

    \(zw = 3e^{i\frac{\pi}{5}} \times 4e^{i\frac{3\pi}{10}} = 12e^{i\left(\frac{\pi}{5} + \frac{3\pi}{10}\right)} = 12e^{i\frac{\pi}{2}}\)

    The numbers in front multiply.

  2. Use Euler's formula

    \(e^{i\theta} = \cos\theta + i\sin\theta,\quad 12e^{i\frac{\pi}{2}} = 12\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)\)

    From the calculator, or from the unit circle: \(\cos\frac{\pi}{2} = 0\) and \(\sin\frac{\pi}{2} = 1\).

  3. Write the answer

    \(zw = 12i\)

    The real part is 0. \(y = 12\).

Video walkthrough coming
The conjugate in exponential form

\(z = 4e^{i\frac{\pi}{3}}\). Write \(z^*\) in exponential form with \(0 \le \theta < 2\pi\), and find \(zz^*\). The argument of \(z^*\) is \(\frac{k\pi}{3}\).

Working
  1. Reflect in the real axis

    \(z^* = 4e^{-i\frac{\pi}{3}}\)

    The conjugate has the same modulus and the opposite argument.

  2. Bring it into range

    \(-\frac{\pi}{3} + 2\pi = \frac{5\pi}{3},\quad z^* = 4e^{i\frac{5\pi}{3}}\)

    A negative argument is the same direction as the one \(2\pi\) larger.

  3. Find zz*

    \(zz^* = 4e^{i\frac{\pi}{3}} \times 4e^{-i\frac{\pi}{3}} = 16e^{0} = 16\)

    The arguments cancel, so the product is real. It is \(|z|^2 = 4^2\).

Video walkthrough coming
Points on a spiral

The points \(z_\theta = \frac{1}{3}\theta e^{i\theta}\), where \(\theta \ge 0\), lie on a curve. (a) Find \(z_\theta\) for \(\theta = \frac{\pi}{2},\ \pi,\ \frac{3\pi}{2}\) and plot the points. (b) Find the value of \(\theta\) for which \(|z_\theta| = 2\), and write that point in the form \(x + iy\).

−3−2−1123−3−2−1123ReIm
Working
  1. Read the modulus and argument

    \(|z_\theta| = \frac{\theta}{3}\) and \(\arg z_\theta = \theta\)

    The number in front is the modulus and the power of \(e\) holds the argument. Both grow together.

  2. The three points

    \(z_{\pi/2} = \frac{\pi}{6}e^{i\frac{\pi}{2}} = 0.524i\)<br>\(z_{\pi} = \frac{\pi}{3}e^{i\pi} = -1.05\)<br>\(z_{3\pi/2} = \frac{\pi}{2}e^{i\frac{3\pi}{2}} = -1.57i\)

    Each point is on an axis, a quarter turn after the last, and further from the origin. Plot \((0, 0.524)\), \((-1.05, 0)\) and \((0, -1.57)\).

  3. Solve for theta

    \(\frac{\theta}{3} = 2 \Rightarrow \theta = 6\)

    The angle is in radians. 6 radians is a little under one full turn.

  4. −3−2−1123−3−2−1123ReImθ = π/2θ = πθ = 3π/2θ = 6
    Cartesian form

    \(z_6 = 2e^{6i} = 2\cos 6 + 2\sin 6\,i = 1.92 - 0.559i\)

    Radian mode. The point is in the fourth quadrant, which agrees with an angle just under \(2\pi\).

Video walkthrough coming