Modulus and Argument of Complex Numbers

The Length and Direction of a Complex Vector

A complex number $z = x + iy$ can be represented as a point $(x, y)$ on the complex plane (similar to the Cartesian plane). Besides being a point, we can also view it as a vector starting from the origin $(0, 0)$ to the point $(x, y)$.

This vector has a length and a direction. This length and direction are what we call the Modulus and Argument.

Modulus of a Complex Number

The Modulus of a complex number $z = x + iy$, written as $|z|$, is the distance from the origin $(0,0)$ to the point $(x, y)$ on the complex plane. This is the same as the length of the vector representing $z$.

To calculate the modulus, we can use the Pythagorean Theorem on the right-angled triangle formed by the real part ($x$), the imaginary part ($y$), and the modulus ($|z|$) as the hypotenuse.

Definition of Modulus:

The modulus of the complex number $z = x + iy$ is:

The modulus is always non-negative (never negative) because it represents a distance.

Calculating the Modulus

The four computations run from a first-quadrant triple through a general pair to the real and imaginary axes, so you see the same square-root formula handle every position.

1. Find the modulus of $z_1 = 3 + 4i$, with $x=3, y=4$

   $$\begin{aligned} |z_1| &= \sqrt{3^2 + 4^2} \\ &= \sqrt{9 + 16} \\ &= \sqrt{25} \\ &= 5 \end{aligned}$$

2. Find the modulus of $z_2 = -1 - 2i$, with $x=-1, y=-2$

   $$\begin{aligned} |z_2| &= \sqrt{(-1)^2 + (-2)^2} \\ &= \sqrt{1 + 4} \\ &= \sqrt{5} \end{aligned}$$

3. Find the modulus of $z_3 = 5$, with $x=5, y=0$

   $$|z_3| = \sqrt{5^2 + 0^2} = \sqrt{25} = 5$$

   (The modulus of a real number is its absolute value).

4. Find the modulus of $z_4 = -2i$, with $x=0, y=-2$

   $$|z_4| = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$$

Argument of a Complex Number

The Argument of a non-zero complex number $z = x + iy$, written as $\arg(z)$ or $\theta$, is the angle formed by the vector $z$ with the positive real axis on the complex plane. This angle is usually measured in radians or degrees.

From basic trigonometry on the same right-angled triangle as in the modulus visualization, we know the relationships:

When $x$ and $y$ are both nonzero, first calculate the positive reference angle:

The ratio $|y/x|$ is positive and finite, so the reference angle is acute: $0^\circ<\alpha<90^\circ$.

The signs of $x$ and $y$ locate the quadrant. They determine how the reference angle becomes the Principal Argument in $(-180^\circ, 180^\circ]$:

• Quadrant $I$ ($x>0, y>0$):

  $$\text{Arg}(z) = \alpha$$

• Quadrant $II$ ($x<0, y>0$):

  $$\text{Arg}(z) = 180^\circ - \alpha$$

• Quadrant $III$ ($x<0, y<0$):

  $$\text{Arg}(z) = -180^\circ + \alpha$$

• Quadrant $IV$ ($x>0, y<0$):

  $$\text{Arg}(z) = -\alpha$$

Points on an axis do not require a quotient $y/x$. Read their direction directly:

Calculating the Argument

For each point, first locate the quadrant, then compute the reference angle $\alpha$ from $|y/x|$, then apply the quadrant rule above.

1. Find the Principal Argument of $z_1 = 1 + i$

   The point $(1, 1)$ is in Quadrant $I$.

   $$\tan \theta = \frac{y}{x} = \frac{1}{1} = 1$$

   $$\begin{aligned} &\theta = \arctan(1) = 45^\circ \text{ or } \frac{\pi}{4} \text{ radians} \end{aligned}$$

2. Find the Principal Argument of $z_2 = -\sqrt{3} + i$

   The point $(-\sqrt{3}, 1)$ is in Quadrant $II$.

   $$\tan \theta = \frac{y}{x} = \frac{1}{-\sqrt{3}}$$

   $$\begin{aligned} \text{Base angle} &= \arctan\left(\left|\frac{1}{-\sqrt{3}}\right|\right) \\ &= \arctan\left(\frac{1}{\sqrt{3}}\right) \\ &= 30^\circ \end{aligned}$$

   $$\begin{aligned} \theta &= 180^\circ - 30^\circ \\ &= 150^\circ \text{ or } \frac{5\pi}{6} \text{ radians} \end{aligned}$$

   Quadrant $II$ lies above the negative real axis, so subtract the reference angle from $180^\circ$.

3. Find the Principal Argument of $z_3 = -1 - i\sqrt{3}$

   The point $(-1, -\sqrt{3})$ is in Quadrant $III$.

   $$\tan \theta = \frac{y}{x} = \frac{-\sqrt{3}}{-1} = \sqrt{3}$$

   $$\text{Base angle} = \arctan(\sqrt{3}) = 60^\circ$$

   $$\begin{aligned} \text{Arg}(z_3) &= -180^\circ + 60^\circ \\ &= -120^\circ \\ &= -\frac{2\pi}{3} \end{aligned}$$

   Quadrant $III$ lies below the negative real axis, so its Principal Argument is negative.

4. Find the Principal Argument of $z_4 = 3 - 3i$

   The point $(3, -3)$ is in Quadrant $IV$.

   $$\tan \theta = \frac{y}{x} = \frac{-3}{3} = -1$$

   $$\begin{aligned} \text{Base angle} &= \arctan(|-1|) \\ &= \arctan(1) \\ &= 45^\circ \end{aligned}$$

   $$\text{Arg}(z_4) = -45^\circ = -\frac{\pi}{4}$$

   Quadrant $IV$ lies below the positive real axis, so its Principal Argument is negative.

Exercise

Find the modulus and Principal Argument (in degrees) of the following complex numbers. For each number, compute the distance from the origin first, then choose the angle inside $(-180^\circ, 180^\circ]$.

1. $z_a = 2 + 2i$
2. $z_b = -4$
3. $z_c = -i$

Worked Solutions

1. For $z_a = 2 + 2i$:

   $x=2, y=2$ (Quadrant $I$) Modulus:

   $$\begin{aligned} |z_a| &= \sqrt{2^2 + 2^2} \\ &= \sqrt{4+4} \\ &= \sqrt{8} \\ &= 2\sqrt{2} \end{aligned}$$

   Argument:

   $$\tan\theta=\frac22=1$$

   The point is in Quadrant $I$, so its Principal Argument is the acute angle:

   $$\theta=\arctan(1)=45^\circ$$

2. For $z_b = -4$:

   $z_b = -4 + 0i$. $x=-4, y=0$ (Negative real axis) Modulus:

   $$|z_b| = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$$

   Argument: The point is on the negative real axis.

   $$\theta = 180^\circ$$

3. For $z_c = -i$:

   $z_c = 0 - 1i$. $x=0, y=-1$ (Negative imaginary axis) Modulus:

   $$|z_c| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1$$

   Argument: The point is on the negative imaginary axis.

   $$\text{Arg}(z_c) = -90^\circ$$