Geometrically, addition of two complex numbers $Z_1$ and $Z_2$ can be visualized as
addition of the vectors by using the *parallelogram law*. The vector sum
$Z_1+Z_2$ is represented by the diagonal of the parallelogram formed by the two
original vectors.

The easiest way to represent the difference $Z_1-Z_2$ is to think in terms of adding a negative vector $Z_1 + \left(-Z_2\right)$. The negative vector is the same vector as its positive counterpart, only pointing in the opposite direction.

Use the following applet to explore this geometric interpretation. Activate the boxes below to show the addition or substraction. You can also drag the points $Z_1$ and $Z_2$ around.

**Exercise 1:** Can you think about a geometric interpretation of the addition
of three complex numbers? In general, what would be a geometric interpretation of the addition of
$n$ complex numbers?

In the previous section we defined the multiplication of two complex numbers $Z_1 $ and $Z_2$ as \begin{eqnarray*} Z_1 Z_2 &=& \left( x_1 + i y_1 \right) \left( x_2 + i y_2 \right)\\ &=& (x_1x_2-y_1y_2) + i(x_1y_2+x_2y_1). \end{eqnarray*} In this case, to appreciate what happens geometrically we need to consider the polar form of $Z_1 $ and $Z_2$. That is \begin{eqnarray*} Z_1 &=& r_1 \left( \cos \phi_1 + i \sin \phi_1 \right) \\ Z_2 &=& r_2 \left( \cos \phi_2 + i \sin \phi_2 \right) \end{eqnarray*} Then the product can be written in the form \begin{eqnarray*} Z_1 Z_2 &=& r_1 r_2 \big[ \left(\cos \phi_1 \cos\phi_2 - \sin \phi_1 \sin \phi_2\right) \big.\\ &+& \big. i\left(\sin \phi_1 \cos\phi_2 + \cos \phi_1 \sin \phi_2\right)\big]. \end{eqnarray*} Now by means of the addition theorems of the sine and cosine this expression can be simplified to \begin{eqnarray*} Z_1 Z_2 &=& r_1 r_2 \big[ \cos \left( \phi_1 +\phi_2 \right) + i \sin \left( \phi_1 +\phi_2 \right)\big]. \end{eqnarray*} Thus the product $Z_1Z_2$ has the modulus $r_1r_2$ and the argument $\phi_1+\phi_2$.

**Exercise 2:**
Consider now
\begin{eqnarray*}
Z_1 &=& r_1 \left( \cos \phi_1 + i \sin \phi_1 \right) \\
Z_2 &=& r_2 \left( \cos \phi_2 + i \sin \phi_2 \right)
\end{eqnarray*}
such that $Z_2\neq 0$. Find the polar representation of $Z_1/Z_2$.
What is the geometric interpretation of this expression?

In the applet below a set of points are defined randomly on the complex plane. Then each point is multiplied by a given complex number $z$. On the right-side screen, drag around the point $z$ and analize the behaviour of the points (⭕) multiplied by $z$ and try to answer the following questions:

- What happens when $z$ is inside, or outside, the unit circle?
- What happens if $z$ moves only around the unit circle?

`Multiply by 1/z`

.

As you already have noticed the geometric interpretation of multiplication of complex numbers is stretching (or squeezing) and rotation of vectors in the plane.

In the previous applet, with the option `Multiply by z`

, set `n = 1`

by dragging the
slider to the left side.
In this case, the applet shows the three complex numbers
$$z_0, z \text{ and } z_1 = z_0\cdot z,$$
represented as vectors.
When $z_0$ and $z$ are non zero, then

- the modulus of $z_1$ is equal to $|z_0 \cdot z|$, and
- the argument of $z_1$ is equal to $\text{Arg }(z_0+z)$.

**Exercise 3:**
Use the same applet, with the option `Multiply by 1/z`

, to investigate
what happens when we multiply by $1/z$.
Set `n = 1`

by dragging the slider to the left side to
show the three complex numbers
$$z_0, z \text{ and } z_2 = z_0\cdot \frac{1}{z}.$$
What happens to the modulus and argument of $z_2$?

NEXT: The Principal Argument