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AI HL · AHL 6

Complex Numbers

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AHL 6.6

Modulus-argument (polar) form

AHL 6.6

Modulus-argument (polar) form

Target: I can find the modulus and argument of a complex number, taking the quadrant into account. (+4 more)

By the end of this section

Key points

Modulus and argument

  • The modulus \(r = |z|\) is the distance from the origin. The argument \(\theta = \arg z\) is the angle from the positive real axis, measured anticlockwise, in radians.
  • For \(z = x + iy\): \(r = \sqrt{x^2 + y^2}\) and \(\tan\theta = \dfrac{y}{x}\).
  • The argument is given in \(0 \le \theta < 2\pi\) or in \(-\pi < \theta \le \pi\). The question says which.
  • Always sketch the point first. \(\tan^{-1}\) gives an angle in the first or fourth quadrant only, so the quadrant decides what to add or subtract.

Finding the argument by quadrant

  • Find the acute angle \(\alpha = \tan^{-1}\left|\dfrac{y}{x}\right|\), using the positive lengths.
  • For \(0 \le \theta < 2\pi\): first quadrant \(\theta = \alpha\), second \(\pi - \alpha\), third \(\pi + \alpha\), fourth \(2\pi - \alpha\).
  • For \(-\pi < \theta \le \pi\): first \(\alpha\), second \(\pi - \alpha\), third \(-(\pi - \alpha)\), fourth \(-\alpha\).
  • The modulus is never negative. \(-2\,\text{cis}\,\theta\) is not in modulus-argument form.

The form r cis theta

  • \(z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta\).
  • To go back to Cartesian form: \(x = r\cos\theta\) and \(y = r\sin\theta\). Use exact values when the angle is a standard one, and radian mode on the calculator otherwise.
  • The conjugate is the reflection in the real axis: if \(z = r\,\text{cis}\,\theta\), then \(z^* = r\,\text{cis}(-\theta)\). Add \(2\pi\) if the angle must be in \(0 \le \theta < 2\pi\).

On the calculator

Converting to polar and back

  • In Radian mode and Complex mode, the complex number tools can give the magnitude \(r\) and the polar angle \(\theta\) of a number, or convert the whole number to polar or to rectangular form.
  • The calculator gives \(\theta\) between \(-\pi\) and \(\pi\). If the question asks for \(0 \le \theta < 2\pi\), add \(2\pi\) to a negative angle.
  • Check the angle against your sketch: the calculator is right, but only your sketch tells you which quadrant you expect.

Screenshot and video to come

Typing a number as r angle theta

  • A number in polar form is typed as \(r\angle\theta\), with the angle symbol from the symbol palette.
  • In Radian mode, \(\theta\) is typed in radians. The calculator returns the number in the form \(x + iy\) in Rectangular format.
  • Check that the calculator is in Radian mode before you start: an angle typed in the wrong mode gives a wrong answer without any warning.

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?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Modulus and argument in the first quadrant

Write \(z = 2 + 5i\) in modulus-argument form. Give \(r\) and \(\theta\) to 3 significant figures, with \(\theta\) in radians.

Working
  1. −1123456−11234567ReImz
    Sketch the point

    \(z = 2 + 5i\) is the point \((2, 5)\)

    It is in the first quadrant, so the angle from the positive real axis is the acute angle.

  2. The modulus

    \(r = \sqrt{2^2 + 5^2} = \sqrt{29} = 5.39\)

    Pythagoras: the distance from the origin.

  3. The argument

    \(\tan\theta = \dfrac{5}{2},\quad \theta = \tan^{-1}\left(\dfrac{5}{2}\right) = 1.19\)

    Radian mode. In the first quadrant, \(\theta\) is the calculator's answer as it stands.

  4. Write the form

    \(z = 5.39\,\text{cis}\,1.19\)

    \(\text{cis}\,\theta\) means \(\cos\theta + i\sin\theta\).

Video walkthrough coming
A number in the third quadrant

Write \(z = -2 - 2\sqrt{3}\,i\) in modulus-argument form, with \(0 \le \theta < 2\pi\). Then give \(\theta\) in the interval \(-\pi < \theta \le \pi\).

Working
  1. −4−3−2−112−5−4−3−2−112ReImz
    Sketch the point

    \(z\) is the point \((-2, -2\sqrt{3})\), in the third quadrant

    Sketch first: the angle you need is the long way round from the positive real axis.

  2. The modulus

    \(r = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4\)

    A whole number: the 4 comes from \(\sqrt{16}\).

  3. The acute angle

    \(\tan\alpha = \dfrac{2\sqrt{3}}{2} = \sqrt{3},\quad \alpha = \dfrac{\pi}{3}\)

    \(\alpha\) is the angle between the vector and the real axis. Use the positive lengths.

  4. Use the quadrant

    \(\theta = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}\), so \(z = 4\,\text{cis}\,\dfrac{4\pi}{3}\)

    Third quadrant: \(\pi + \alpha\).

  5. The other interval

    \(\theta = \dfrac{4\pi}{3} - 2\pi = -\dfrac{2\pi}{3}\)

    Take away \(2\pi\) to get into \(-\pi < \theta \le \pi\). It is the same point.

Video walkthrough coming
Converting with the calculator

Use the calculator to write \(z = -4 + 4i\) in modulus-argument form, with \(\theta\) to 3 significant figures. Then give the exact form.

Working
  1. Sketch the point

    \((-4, 4)\) is in the second quadrant

    The acute angle with the negative real axis is \(\dfrac{\pi}{4}\), so \(\theta = \pi - \dfrac{\pi}{4}\).

  2. Convert to polar

    From the calculator (Convert to Polar, radian mode): \(5.66\angle 2.36\)

    The calculator gives \(r\) and \(\theta\), with \(\theta\) between \(-\pi\) and \(\pi\). Check that the angle is in the quadrant you sketched.

  3. Exact form

    \(r = \sqrt{16 + 16} = 4\sqrt{2},\quad \theta = \dfrac{3\pi}{4}\)

    \(z = 4\sqrt{2}\,\text{cis}\,\dfrac{3\pi}{4}\).

Video walkthrough coming
Cartesian form from r and theta, exact

A complex number \(z\) has modulus 6 and argument \(\dfrac{2\pi}{3}\). Write it in the form \(x + iy\).

Working
  1. The formulae

    \(x = r\cos\theta,\quad y = r\sin\theta\)

    These come from the right-angled triangle with hypotenuse \(r\).

  2. The real part

    \(x = 6\cos\dfrac{2\pi}{3} = 6\left(-\dfrac{1}{2}\right) = -3\)

    Use the exact value, not a decimal.

  3. The imaginary part

    \(y = 6\sin\dfrac{2\pi}{3} = 6\left(\dfrac{\sqrt{3}}{2}\right) = 3\sqrt{3}\)

    \(\sin\) is positive in the second quadrant.

  4. State it

    \(z = -3 + 3\sqrt{3}\,i\)

    Check the quadrant: negative real part, positive imaginary part: second quadrant.

Video walkthrough coming
Cartesian form from r and theta, decimal

A complex number has modulus 10 and argument 0.9 radians. Write it in the form \(x + iy\), to 3 significant figures.

Working
  1. Set the mode

    Calculator in radian mode

    A common slip is to leave it in degrees.

  2. The real part

    From the calculator: \(x = 10\cos 0.9 = 6.22\)

    \(r\cos\theta\).

  3. The imaginary part

    From the calculator: \(y = 10\sin 0.9 = 7.83\)

    \(r\sin\theta\).

  4. State it

    \(z = 6.22 + 7.83i\)

    Check: \(6.22^2 + 7.83^2 \approx 100\), so the modulus is 10.

Video walkthrough coming
The conjugate in polar form

\(z = 3\,\text{cis}\,\dfrac{5\pi}{4}\). Write \(z^*\) in modulus-argument form with \(-\pi < \theta \le \pi\).

Working
  1. −4−3−2−112−4−3−2−11234ReImzz*
    Reflect in the real axis

    \(z^* = 3\,\text{cis}\left(-\dfrac{5\pi}{4}\right)\)

    The modulus does not change, and the angle changes sign: if \(z = r\,\text{cis}\,\theta\), then \(z^* = r\,\text{cis}(-\theta)\).

  2. Bring the angle into range

    \(-\dfrac{5\pi}{4} + 2\pi = \dfrac{3\pi}{4}\)

    Add \(2\pi\) to a negative angle below \(-\pi\).

  3. State it

    \(z^* = 3\,\text{cis}\,\dfrac{3\pi}{4}\)

    Check on the diagram: \(z^*\) is in the second quadrant, at \(\dfrac{3\pi}{4}\) from the positive real axis.

Video walkthrough coming