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AI HL · AHL 6

Complex Numbers

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AHL 6.2

Conjugates, division and equating parts

AHL 6.2

Conjugates, division and equating parts

Target: I can find the complex conjugate of a number and use \(zz^* = x^2 + y^2\). (+4 more)

By the end of this section

Key points

The conjugate

  • If \(z = x + iy\), the conjugate is \(z^* = x - iy\): change the sign of the imaginary part only.
  • \(zz^* = x^2 + y^2\). It is always real and never negative. The two middle terms cancel and \(-i^2y^2 = +y^2\).
  • \(z + z^* = 2x\) is real, and \(z - z^* = 2iy\) is purely imaginary.

Dividing complex numbers

  • To divide, multiply the top and the bottom by the conjugate of the denominator. This makes the denominator real, as rationalising does for a surd.
  • Expand the top, and use \(x^2 + y^2\) for the bottom. Then divide each part of the top by the real denominator.
  • Leave exact fractions unless the question asks for decimals. Do not leave an \(i\) in the denominator.

Equating real and imaginary parts

  • Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal. One complex equation gives two real equations.
  • To solve an equation in \(z\) and \(z^*\), write \(z = x + iy\) and \(z^* = x - iy\), expand, group the real and imaginary parts, then equate.
  • The same method finds real unknowns \(a\) and \(b\) in a product: expand the product, group, equate, solve the two equations together.
  • Check by substituting your answer back into the original equation.

On the calculator

Checking conjugates and division

  • In Complex mode, type a quotient. The calculator returns it in the form \(x + iy\).
  • Use it to check your division, and to find a conjugate when you need to check \(z^*\).
  • Your hand working is what is marked. The calculator checks it.

Screenshot and video to come

Solving the two equations

  • After equating real and imaginary parts you have two linear equations in two unknowns.
  • Solve them as a pair of simultaneous equations on the calculator.
  • Write both equations in your answer, then the solution.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
The conjugate and zz*

\(z = 5 - 12i\). Write down \(z^*\) and find \(zz^*\).

Working
  1. Change the sign of the imaginary part

    \(z^* = 5 + 12i\)

    The conjugate has the same real part and the opposite imaginary part.

  2. Multiply

    \(zz^* = (5 - 12i)(5 + 12i) = 25 + 60i - 60i - 144i^2\)

    The two middle terms cancel.

  3. Use i² = −1

    \(zz^* = 25 + 144 = 169\)

    The result is real. Check: \(x^2 + y^2 = 5^2 + 12^2\).

Video walkthrough coming
Showing results about z and z*

Let \(z = x + iy\) with \(x\) and \(y\) real. Show that \(zz^*\) and \(z + z^*\) are real. Then, for \(z = 7 + 4i\), write \(z - z^*\) in the form \(ki\) and give \(k\).

Working
  1. The product

    \(zz^* = (x + iy)(x - iy) = x^2 - ixy + ixy - i^2y^2 = x^2 + y^2\)

    \(i^2 = -1\). \(x\) and \(y\) are real, so \(x^2 + y^2\) is real.

  2. The sum

    \(z + z^* = (x + iy) + (x - iy) = 2x\)

    The imaginary parts cancel. \(2x\) is real.

  3. The difference

    \(z - z^* = (x + iy) - (x - iy) = 2iy\)

    The real parts cancel. The result is a pure imaginary number.

  4. Put in the numbers

    \(z - z^* = 2i \times 4 = 8i, \text{ so } k = 8\)

    \(y = 4\).

Video walkthrough coming
Dividing by a complex number

\(z = 12 - 5i\) and \(w = 3 + 2i\). Find \(\dfrac{z}{w}\) in the form \(x + iy\).

Working
  1. Multiply top and bottom by the conjugate of w

    \(\dfrac{12 - 5i}{3 + 2i} \times \dfrac{3 - 2i}{3 - 2i}\)

    This is like rationalising a denominator. The conjugate of \(3 + 2i\) is \(3 - 2i\).

  2. Top

    \((12 - 5i)(3 - 2i) = 36 - 24i - 15i + 10i^2 = 26 - 39i\)

    Expand and use \(i^2 = -1\).

  3. Bottom

    \((3 + 2i)(3 - 2i) = 3^2 + 2^2 = 13\)

    A conjugate pair always gives \(x^2 + y^2\), a real number.

  4. Divide

    \(\dfrac{26 - 39i}{13} = 2 - 3i\)

    Divide each part by 13.

  5. Check

    From the calculator (complex mode): \(2 - 3i\)

    Type the quotient and compare.

Video walkthrough coming
A division that does not simplify

Find \(\dfrac{1 + 4i}{2 + 3i}\) in the form \(x + iy\), with \(x\) and \(y\) as exact fractions. Give \(x\) to 3 significant figures.

Working
  1. Multiply by the conjugate

    \(\dfrac{(1 + 4i)(2 - 3i)}{(2 + 3i)(2 - 3i)}\)

    The denominator has conjugate \(2 - 3i\).

  2. Top

    \((1 + 4i)(2 - 3i) = 2 - 3i + 8i - 12i^2 = 14 + 5i\)

    \(-12i^2 = +12\).

  3. Bottom and answer

    \(\dfrac{14 + 5i}{13} = \dfrac{14}{13} + \dfrac{5}{13}i\)

    Not every quotient is a whole number. Leave exact fractions, or use \(1.08 + 0.385i\).

Video walkthrough coming
An equation in z and z*

Find the complex number \(z\) such that \(2z - 3z^* = 10 + 15i\).

Working
  1. Write z and z*

    \(z = x + iy,\ z^* = x - iy\)

    Replace both, then expand.

  2. Expand and group

    \(2(x + iy) - 3(x - iy) = -x + 5iy\)

    \(2x - 3x = -x\) and \(2iy + 3iy = 5iy\).

  3. Equate real parts and imaginary parts

    \(-x = 10,\quad 5y = 15\)

    Two complex numbers are equal only when both parts are equal.

  4. Solve

    \(x = -10,\ y = 3,\ z = -10 + 3i\)

    State the answer as a complex number.

Video walkthrough coming
Finding real a and b

\(a\) and \(b\) are real numbers. \((a + bi)(2 - i) = 1 - 8i\). Find \(a\) and \(b\).

Working
  1. Expand

    \((a + bi)(2 - i) = 2a - ai + 2bi - bi^2 = (2a + b) + (2b - a)i\)

    \(-bi^2 = +b\). Group real and imaginary parts.

  2. Equate real and imaginary parts

    \(2a + b = 1,\quad 2b - a = -8\)

    Two equations in two real unknowns.

  3. Solve

    From the calculator (simultaneous equations): \(a = 2,\ b = -3\)

    Or substitute \(b = 1 - 2a\) into the second equation.

  4. Check

    \((2 - 3i)(2 - i) = 4 - 2i - 6i + 3i^2 = 1 - 8i\)

    Substitute back.

Video walkthrough coming