Modulus, argument and loci

Length and angle

For z = x + iy ≠ 0, |z| = √(x² + y²). In this course the principal argument is taken in (−π, π]; other sources may use a different principal interval. Plot the point to determine the quadrant. The origin has modulus 0 and no argument. Modulus and argument together determine the point, and the conversion is the trigonometry of one right-angled triangle:

with r = |z| and θ = arg z. Check the current formula booklet for your awarding organisation; whether or not the result is supplied, you must know when it applies and how to derive it if the question asks for a proof. The calculator's tan⁻¹(y/x) returns a value in (−π/2, π/2), so for a number such as −1 + i it gives −π/4, which must be corrected to 3π/4 for a point in the second quadrant. Plot the point and adjust the inverse tangent accordingly.

Two shorthands for the same form are used by some awarding organisations, including OCR, and should be recognised on sight. One is the square bracket [r, θ], the modulus and the argument in that order. The other is r cis θ, in which cis abbreviates cos + i sin. Each denotes r(cos θ + i sin θ), so 2 cis(π/3) and [2, π/3] both denote the number 1 + √3 i in the figure below. Convert either into the form the working needs.

Multiplying in modulus-argument form

Multiplication has a direct geometric description: moduli multiply and arguments add. Division divides the moduli and subtracts the arguments. A product that is laborious in x + yi form is often short in modulus-argument form.

Two conditions apply. Neither number may be zero, since zero has no argument; and the sum or difference of two arguments may lie outside the principal interval −π < θ ≤ π, in which case add or subtract 2π to return it to that interval before quoting it. In general the rule for arguments holds only modulo 2π.

After adding two arguments the result may lie outside the principal interval, so subtract 2π to return it to (−π, π] before quoting it.

Loci on the Argand diagram

An equation in z defines a set of points, a locus, and three forms occur in this course. |z − a| = r says 'distance from a is r', a circle of centre a and radius r. |z − a| = |z − b| says 'equidistant from a and b', giving the perpendicular bisector of the segment joining them. arg(z − a) = θ says 'the direction from a is θ', a half-line from a at angle θ, with a itself excluded, since z = a would ask for the argument of zero.

Regions defined by inequalities

Replacing the equals sign by an inequality gives a region. |z − a| < r is the interior of the circle and |z − a| > r the exterior. |z − a| ≤ |z − b| is the half plane containing a, bounded by the perpendicular bisector, with the boundary included. Where two conditions are combined, draw each boundary first, decide which side is required by testing one convenient point such as the origin, then shade the intersection. Draw strict boundaries dashed and inclusive boundaries solid. A region defined by an argument condition excludes the point at which the argument is undefined.

Describe the drawing

An Argand diagram carrying the dashed circle of the strict inequality, centre one plus two i and radius two point five, and the solid perpendicular bisector of one plus two i and five, with the overlap of the open disc and the half plane shaded: the origin, ringed as the test point, is two point two four from a, closer than the radius and closer than it is to b, so it sits inside the shaded region.