Mathematics 9709 · Pure Mathematics 3

The Argand diagram: modulus and argument

Plot $z=x+iy$ as the point $(x,y)$ on the Argand diagram. It becomes an arrow from the origin O.
Two numbers fix that arrow: the modulus $|z|$ (how long it is) and the argument $\arg z$ (the angle it makes with the positive real axis).
These are just the polar coordinates of the same point.

$|z|=\sqrt{x^2+y^2}$ is the length; $\arg z$ is the angle. To save marks, sketch it first. The quadrant your point lands in, not your calculator, tells you the sign and size of the argument.

Modulus = length, argument = angle

$|z|=\sqrt{x^2+y^2},\qquad \tan(\arg z)=\tfrac{y}{x}\ \ (-\pi<\arg z\le\pi)$

The principal argument stays in $(-\pi,\pi]$ : anticlockwise is positive, clockwise is negative.
Your calculator's $\tan^{-1}\tfrac{y}{x}$ only gives the acute reference angle off the real axis. The quadrant tells you what to do with it:

1st quadrant ( $x>0,\,y>0$ ): $\arg z=\alpha$ directly.
2nd ( $x<0,\,y>0$ ): $\arg z=\pi-\alpha$ .
3rd ( $x<0,\,y<0$ ): $\arg z=-\pi+\alpha$ .
4th ( $x>0,\,y<0$ ): $\arg z=-\alpha$ .

where $\alpha=\tan^{-1}\left|\tfrac{y}{x}\right|$ is the acute angle.
Drag $z$ in the picture below. The argument flips sign as $z$ crosses below the real axis, and it jumps near $\pm\pi$ along the negative real axis.

Re zIm z

show locus |z − a| ≤ 2.5, centre a = −2 + i

z = 3.00 + 2.00i·|z| = 3.61·arg z = 0.59 rad

|z| is the length of Oz (here 5); arg z is the angle it makes with the positive real axis (here about 0.93 rad).

Why the principal argument uses (−π, π], not 0 to 2π

You can add $2\pi$ to any angle and get the same direction, so we need one agreed answer. The syllabus picks $-\pi<\arg z\le\pi$ : measure the shorter way round, anticlockwise positive and clockwise negative. This keeps points just below the positive real axis at a small negative argument (not nearly $2\pi$ ). That is why a sketch tells you straight away whether the answer is positive or negative.

Worked examples

Worked example

Modulus and argument of $-3+4i$

Textbook Worked Example 11.5(b): a point in the second quadrant.

Length: $|z|=\sqrt{(-3)^2+4^2}=\sqrt{25}=5$ .
Reference angle: $\alpha=\tan^{-1}\tfrac43=0.927$ . The point sits in the second quadrant (left and up).
Second quadrant means $\arg z=\pi-\alpha=\pi-0.927$ .

Answer

|z|=5,\quad \arg z=2.21\ \text{rad}

Worked example

Write $3\sqrt2\big(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\big)$ as $x+iy$

Real part: $x=3\sqrt2\cos\tfrac{\pi}{4}=3\sqrt2\cdot\tfrac{1}{\sqrt2}=3$ .
Imaginary part: $y=3\sqrt2\sin\tfrac{\pi}{4}=3$ .

Answer

3+3i

Where the marks go

Common mistake

Keep the principal argument in

(-\pi,\pi]

, and let the quadrant decide — not just

\tan^{-1}\tfrac{y}{x}

, which a calculator always gives between

\pm\tfrac{\pi}{2}

. For a negative real like

-5

\arg=\pi

, not

0

Common mistake

|z|

is never negative — it is a length. If your working gives a negative modulus, you dropped a sign before the square root, not after.

Now you try

Find $|z|$ and $\arg z$ for $z=-4-3i$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Find the length, then fix the angle by the quadrant: $|z|=\sqrt{16+9}=5$ ; third quadrant, so $\arg z=-\pi+\tan^{-1}\tfrac34$ :

Answer

|z|=5,\quad \arg z=-2.50\ \text{rad}

PreviousComplex arithmetic and the conjugate NextPolar and exponential form; multiply and divide

The Argand diagram: modulus and argument

Plot $z=x+iy$ as the point $(x,y)$ on the Argand diagram. It becomes an arrow from the origin O.
Two numbers fix that arrow: the modulus $|z|$ (how long it is) and the argument $\arg z$ (the angle it makes with the positive real axis).
These are just the polar coordinates of the same point.

$|z|=\sqrt{x^2+y^2}$ is the length; $\arg z$ is the angle. To save marks, sketch it first. The quadrant your point lands in, not your calculator, tells you the sign and size of the argument.

Modulus = length, argument = angle

$|z|=\sqrt{x^2+y^2},\qquad \tan(\arg z)=\tfrac{y}{x}\ \ (-\pi<\arg z\le\pi)$

The principal argument stays in $(-\pi,\pi]$ : anticlockwise is positive, clockwise is negative.
Your calculator's $\tan^{-1}\tfrac{y}{x}$ only gives the acute reference angle off the real axis. The quadrant tells you what to do with it:

1st quadrant ( $x>0,\,y>0$ ): $\arg z=\alpha$ directly.
2nd ( $x<0,\,y>0$ ): $\arg z=\pi-\alpha$ .
3rd ( $x<0,\,y<0$ ): $\arg z=-\pi+\alpha$ .
4th ( $x>0,\,y<0$ ): $\arg z=-\alpha$ .

where $\alpha=\tan^{-1}\left|\tfrac{y}{x}\right|$ is the acute angle.
Drag $z$ in the picture below. The argument flips sign as $z$ crosses below the real axis, and it jumps near $\pm\pi$ along the negative real axis.

Re zIm z

show locus |z − a| ≤ 2.5, centre a = −2 + i

z = 3.00 + 2.00i·|z| = 3.61·arg z = 0.59 rad

|z| is the length of Oz (here 5); arg z is the angle it makes with the positive real axis (here about 0.93 rad).

Why the principal argument uses (−π, π], not 0 to 2π

Worked examples

Worked example

Modulus and argument of $-3+4i$

Textbook Worked Example 11.5(b): a point in the second quadrant.

Length: $|z|=\sqrt{(-3)^2+4^2}=\sqrt{25}=5$ .
Reference angle: $\alpha=\tan^{-1}\tfrac43=0.927$ . The point sits in the second quadrant (left and up).
Second quadrant means $\arg z=\pi-\alpha=\pi-0.927$ .

Answer

|z|=5,\quad \arg z=2.21\ \text{rad}

Worked example

Write $3\sqrt2\big(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\big)$ as $x+iy$

Real part: $x=3\sqrt2\cos\tfrac{\pi}{4}=3\sqrt2\cdot\tfrac{1}{\sqrt2}=3$ .
Imaginary part: $y=3\sqrt2\sin\tfrac{\pi}{4}=3$ .

Answer

3+3i

Where the marks go

Common mistake

Keep the principal argument in

(-\pi,\pi]

, and let the quadrant decide — not just

\tan^{-1}\tfrac{y}{x}

, which a calculator always gives between

\pm\tfrac{\pi}{2}

. For a negative real like

-5

\arg=\pi

, not

0

Common mistake

|z|

is never negative — it is a length. If your working gives a negative modulus, you dropped a sign before the square root, not after.

Now you try

Find $|z|$ and $\arg z$ for $z=-4-3i$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Find the length, then fix the angle by the quadrant: $|z|=\sqrt{16+9}=5$ ; third quadrant, so $\arg z=-\pi+\tan^{-1}\tfrac34$ :

Answer

|z|=5,\quad \arg z=-2.50\ \text{rad}