Modulus & Argument (Cambridge International A-Level Mathematics)

Modulus & argument

How do I find the modulus of a complex number?

• The modulus of a complex number is its distance from the origin when plotted on an Argand diagram

• The modulus of $z$ is written open vertical bar z close vertical bar

• If $z=x+iy$, then we can use Pythagoras to show…

  • $\left| z \right| = \sqrt{x^{2} + y^{2}}$

• A modulus is always positive

• the modulus is related to the complex conjugate by…

  • $z z^{asterisk} \times = z^{asterisk} \times z = \left| z \right|^{2}$
  • This is because $z z^{asterisk} \times = ( x + \mathrm{i} y ) ( x - \mathrm{i} y ) = x^{2} + y^{2}$

• In general, $\left| z_{1} + z_{2} \right| \text{eq} \left| z_{1} \right| + \vert z_{2} \vert$

  • e.g. both $z_{1}=3+4i$ and $z_{2}=-3+4i$ have a modulus of 5, but $z_{1}+z_{2}$ simplifies to $8i$ which has a modulus of 8

How do I find the argument of a complex number?

• The argument of a complex number is the anti-clockwise angle that it makes when starting at the positive real axis on an Argand diagram

• Arguments are measured in radians

  • Sometimes these can be given exact in terms of $π$

• The argument of $z$ is written $\mathrm{arg}z$

• Arguments can be calculated using right-angled trigonometry

  • This involves using the tan ratio plus a sketch to decide whether it is positive/negative and acute/obtuse

• Arguments are usually given in the range $-π<\mathrm{arg}z\leqπ$

  • Negative arguments are for complex numbers in the third and fourth quadrants
  • Occasionally you could be asked to give arguments in the range $0\leq\mathrm{arg}z<2π$

• The argument of zero, $\mathrm{arg}0$ is undefined (no angle can be drawn)

Worked Example

Exam Hint

• Give non-exact arguments in radians to 3 significant figures.

Modulus-Argument (Polar) Form

The complex number $z=x+iy$ is said to be in Cartesian form. There are, however, other ways to write a complex number, such as in modulus-argument (polar) form.

How do I write a complex number in modulus-argument (polar) form?

• The Cartesian form of a complex number, $z=x+iy$, is written in terms of its real part, $x$, and its imaginary part, $y$

• If we let $r = \vert z \vert$ and $θ=\mathrm{arg}z$, then it is possible to write a complex number in terms of its modulus, $r$, and its argument, $θ$, called the modulus-argument (polar) form, given by...

  • $z = r ( \cos \\theta + \text{isin} \\theta )$

• It is usual to give arguments in the range $-π<θ\leqπ$

  • Negative arguments should be shown clearly, e.g. $z = 2 ( \cos ( - \frac{\pi }{3} ) + \text{isin} ( - \frac{\pi }{3} ) )$without simplifying $\cos ( - \frac{\pi }{3} )$ to either $\cos ( \frac{\pi }{3} )$ or $\frac{1}{2}$
  • Occasionally you could be asked to give arguments in the range $0\leqθ<2π$

• If a complex number is given in the form $z = r ( \cos \\theta - \text{isin} \\theta )$, then it is not currently in modulus-argument (polar) form due to the minus sign, but can be converted as follows…

  • By considering transformations of trigonometric functions, we see that $- \sin \\theta \equiv \sin ( - \\theta )$ and $\cos \\theta \equiv \cos ( - \\theta )$
  • Therefore $z = r ( \cos \\theta - \text{isin} \\theta )$ can be written as $z = r ( \cos ( - \\theta ) + \text{isin} ( - \\theta ) )$, now in the correct form and indicating an argument of $-θ$

• To convert from modulus-argument (polar) form back to Cartesian form, evaluate the real and imaginary parts

  • E.g. $z = 2 ( \cos ( - \frac{\pi }{3} ) + \text{isin} ( - \frac{\pi }{3} ) )$ becomes $z = 2 ( 1 \frac{1}{2} + \mathrm{i} ( - \frac{ \sqrt{3} }{2} ) ) = 1 - \sqrt{3} \mathrm{i}$

What are the rules for moduli and arguments under multiplication and division?

• When two complex numbers, $z_{1}$ and $z_{2}$, are multiplied to give $z_{1}z_{2}$, their moduli are also multiplied

  • $\left| z_{1} z_{2} \right| = \left| z_{1} \right| \vert z_{2} \vert$

• When two complex numbers, $z_{1}$ and $z_{2}$, are divided to give $\frac{z_{1}}{z_{2}}$, their moduli are also divided

  • | z _ 1 / z _ 2 | = (| z _ 1 |)/(| z _ 2 |)

• When two complex numbers, $z_{1}$ and $z_{2}$, are multiplied to give $z_{1}z_{2}$, their arguments are added

  • $\text{arg} ( z_{1} z_{2} ) = \text{arg} z_{1} + \text{arg} z_{2}$

• When two complex numbers, $z_{1}$and $z_{2}$, are divided to give $\frac{z_{1}}{z_{2}}$, their arguments are subtracted

  • $\text{arg} ( \frac{z_{1}}{z_{2}} ) = \text{arg} z_{1} - \text{arg} z_{2}$

How do I multiply complex numbers in modulus-argument (polar) form?

• The main benefit of writing complex numbers in modulus-argument (polar) form is that they multiply and divide very easily (often quicker than when in Cartesian form)

• To multiply two complex numbers, $z_{1}$ and $z_{2}$, in modulus-argument (polar) form we use the rules from above to multiply their moduli and add their arguments

  • open vertical bar z_ 1z_ 2 close vertical bar=open vertical bar z_ 1 close vertical bar open vertical bar z_ 2 close vertical bar
  • $\mathrm{arg}(z_{1}z_{2})=\mathrm{arg}z_{1}+\mathrm{arg}z_{2}$

• So if $z_{1}=r_{1}(\mathrm{cos}θ_{1}+\mathrm{isin}θ_{1})$ and $z_{2}=r_{2}(\mathrm{cos}θ_{2}+\mathrm{isin}θ_{2})$ then the rules above give…

  • $z_{1} z_{2} = r_{1} r_{2} ( \cos ( \theta_{1} + \theta_{2} ) + \text{isin} ( \theta_{1} + \theta_{2} ) )$

• Sometimes the new argument, $θ_{1}+θ_{2}$, does not lie in the range $-π<θ\leqπ$ (or $0\leqθ<2π$ if this is being used)

  • An out-of-range argument can be adjusted by either adding or subtracting$2π$

  • E.g. If $θ_{1}=\frac{2π}{3}$ and $θ_{2}=\frac{π}{2}$ then $θ_{1}+θ_{2}=\frac{7π}{6}$

    • This is currently not in the range , but by subtracting $2π$ from $\frac{7π}{6}$ to give $-\frac{5π}{6}$, a new argument is formed that lies in the correct range and represents the same angle on an Argand diagram

• The rules of multiplying the moduli and adding the arguments can also be applied when…

  • …multiplying three complex numbers together, $z_{1}z_{2}z_{3}$, or more
  • …finding powers of a complex number (e.g. $z^{2}$ can be written as $zz$)

• Whilst not examinable, the rules for multiplication can be proved algebraically by multiplying $z_{1}=r_{1}(\mathrm{cos}θ_{1}+\mathrm{isin}θ_{1})$ by $z_{2}=r_{2}(\mathrm{cos}θ_{2}+\mathrm{isin}θ_{2})$, expanding the brackets and using compound angle formulae

How do I divide complex numbers in modulus-argument (polar) form?

• To divide two complex numbers, $z_{1}$ and $z_{2}$ in modulus-argument (polar) form, we use the rules from above to divide their moduli and subtract their arguments

  • | z _ 1 / z _ 2 | _ = (| z _ 1 |)/(| z _ 2 |)
  • $\text{arg} ( \frac{z_{1}}{z_{2}} ) = \text{arg} z_{1} - \text{arg} z_{2}$

• So if $z_{1}=r_{1}(\mathrm{cos}θ_{1}+\mathrm{isin}θ_{1})$ and $z_{2}=r_{2}(\mathrm{cos}θ_{2}+\mathrm{isin}θ_{2})$ then the rules above give…

  • z_ 1 / z_ 2=r_ 1 / r_ 2 (cos (θ _ 1 - θ _ 2)+isin (θ _ 1 - θ _ 2))_

• As with multiplication, sometimes the new argument, $θ_{1}-θ_{2}$, can lie out of the range $-π<θ\leqπ$ (or the range $0\leqθ<2π$ if this is being used)

  • You can add or subtract$2π$ to bring out-of-range arguments back in range

• Whilst not examinable, the rules for division can be proved algebraically by dividing $z_{1}=r_{1}(\mathrm{cos}θ_{1}+\mathrm{isin}θ_{1})$ by $z_{2}=r_{2}(\mathrm{cos}θ_{2}+\mathrm{isin}θ_{2})$, using complex division and compound angle formulae

Worked Example

Exam Hint

• The rules for multiplying and dividing in modulus-argument (polar) form must be learnt (they are not given in the formula booklet).
• Remember to add or subtract $2π$ to any out-of-range arguments to bring them back in range.
• If a question does not give a clear range for arguments, then both $-π<θ\leqπ$ or $0<θ\leq2π$ would be accepted.