Measure modulus and argument

Measure modulus and principal argument

Plot the point first. Find its distance from the origin, identify its axis or quadrant, and only then select the angle in the stated principal interval. Zero has no direction to select.

For $z=x+iy$, plot $P(x,y)$ on the Argand plane.

The position vector $\overrightarrow{OP}$ has length $r$ and direction $\theta$ from the positive real axis.

The source triangle labels the horizontal component $x$, vertical component $y$, hypotenuse $r$ and angle $\theta$; the point/vector/direction figure above retains these roles.

The modulus is the distance from $O(0,0)$ to $P(x,y)$:

$$|z|=r=\sqrt{x^2+y^2}.$$

An argument describes the direction of this vector.

Anticlockwise angles are positive and clockwise angles negative.

Arguments differ by multiples of $2\pi$; the notes choose the principal argument in $(-\pi,\pi]$.

In particular the negative real axis uses $\pi$, not $-\pi$.

The source procedure is: draw an Argand diagram and identify the quadrant; find the basic angle $\alpha=\tan^{-1}\lvert y/x\rvert$ for $x\ne0$; then choose the correct principal angle from the cases below.

Treat points on the axes separately.

Principal argument $\theta$

Case: 1

• Position of a nonzero point: Positive real axis
• Principal argument: $0$

Case: 2

• Position of a nonzero point: Negative real axis
• Principal argument: $\pi$

Case: 3

• Position of a nonzero point: Positive imaginary axis
• Principal argument: $\pi/2$

Case: 4

• Position of a nonzero point: Negative imaginary axis
• Principal argument: $-\pi/2$

Case: 5

• Position of a nonzero point: First quadrant
• Principal argument: $\alpha=\tan^{-1}\lvert y/x\rvert$

Case: 6

• Position of a nonzero point: Second quadrant
• Principal argument: $\pi-\alpha=\pi-\tan^{-1}\lvert y/x\rvert$

Case: 7

• Position of a nonzero point: Third quadrant
• Principal argument: $-\pi+\alpha=-\pi+\tan^{-1}\lvert y/x\rvert$

Case: 8

• Position of a nonzero point: Fourth quadrant
• Principal argument: $-\alpha=-\tan^{-1}\lvert y/x\rvert$

The eight diagrams show the same cases, with signed angle arrows from the positive real axis and basic-angle arcs in the quadrants. Their points illustrate the position cases rather than prescribing particular numerical coordinates.

The argument of zero is undefined. The slope expression would require the undefined ratio $0/0$, and geometrically the zero vector has no direction from which to measure an angle.

Its modulus is still $|0|=0$.

Apply the modulus and argument rules with their conditions

Appendix C is on PDF/printed 38–39 in both copies.

The earlier calculator note points to page 42; that printed pointer is stale.

The appendix is labelled non-examinable in the earlier note; retain that source label while using these rules to understand the worked examples.

For z=x+iy with real x,y, and for the products/quotients below where defined, the nine source rows are:

Row: (a)

Rule, with the necessary conditions:

$$zz^*=\lvert z\rvert^2=x^2+y^2$$

Row: (b)

Rule, with the necessary conditions:

$$\lvert z^*\rvert=\lvert z\rvert$$

Row: (c)

Rule, with the necessary conditions:

For nonzero z:

$$\arg(z^*)\equiv-\arg z\pmod{2\pi}$$

Choose the principal representative.

Row: (d)

Rule, with the necessary conditions:

$$\lvert z_1z_2\rvert=\lvert z_1\rvert\lvert z_2\rvert$$

Row: (e)

Rule, with the necessary conditions:

$$\lvert z^n\rvert=\lvert z\rvert^n$$

For positive integer n; negative integer n needs z≠0.

Row: (f)

Rule, with the necessary conditions:

$$\lvert z_1/z_2\rvert=\lvert z_1\rvert/\lvert z_2\rvert$$

With z₂≠0.

Row: (g)

Rule, with the necessary conditions:

$$\arg(z_1z_2)\equiv\arg z_1+\arg z_2\pmod{2\pi}$$

With both numbers nonzero.

Row: (h)

Rule, with the necessary conditions:

$$\arg(z^n)\equiv n\arg z\pmod{2\pi}$$

With nonzero z and integer n.

Row: (i)

Rule, with the necessary conditions:

$$\arg(z_1/z_2)\equiv\arg z_1-\arg z_2\pmod{2\pi}$$

With both numbers nonzero.

The source prints the argument rows as equalities without branch conditions.

For principal arguments in $(-\pi,\pi]$, add or subtract multiples of 2π until the result lies in that interval.

For example, both −1+i and −1−i have argument magnitude 3π/4.

Squaring −1+i gives −2i: the raw sum 3π/2 must become −π/2.

The quotient $(-1+i)/(-1-i)=-i$ gives the same adjustment after subtracting arguments.

The power $(1+i)^6=-8i$ likewise has principal argument −π/2, not 3π/2.

For z=−1, conjugation leaves z unchanged, so both principal arguments are π; the unadjusted −π is excluded.

Zero has modulus 0 but no argument.

The nine worked examples use 1+i and √3+i.

Their results are 2,√2,−π/4,2√2,4,√2/2,5π/12,π andπ/12 in source order.

In these particular examples all raw angles are already principal.

Open the linked appendix exercise for every printed calculation and the relevant condition at each step.