# 3.5: Representing Complex Numbers in Vector Space

So far we have coded the complex number $$z=x+jy$$ with the Cartesian pair $$(x,y)$$ and with the polar pair $$(r∠θ)$$. We now show how the complex number $$z$$ may be coded with a two-dimensional vector $$z$$ and show how this new code may be used to gain insight about complex numbers.

## Coding a Complex Number as a Vector

We code the complex number $$z=x+jy$$ with the two-dimensional vector $$z=[xy]$$:

$x + jy = z ⇔ z = [xy]$

We plot this vector as in Figure. We say that the vector z belongs to a “vector space.” This means that vectors may be added and scaled according to the rules

$z_1+z_2=[x_1+x_2y_1+y_2]$

$az=\begin{bmatrix}ax\\ay\end{bmatrix}$

Furthermore, it means that an additive inverse −z, an additive identity 0, and a multiplicative identity 1 all exist

$z+(−z)=0$

$lz=z$

The vector 0 is $$0=\begin{bmatrix}0\\0\end{bmatrix}$$

Prove that vector addition and scalar multiplication satisfy these properties of commutation, association, and distribution:

$z_1+z_2 = z_2+z_1$

$(z_1+z_2)+z_3=z_1+(z_2+z_3)$

$a(bz)=(ab)z$

$a(z_1+z_2)=az_1+az_2$

## Inner Product and Norm

The inner product between two vectors z1 and z2 is defined to be the real number

$(z_1,z_2)=x_1x_2+y_1y_2$

We sometimes write this inner product as the vector product (more on this in Linear Algebra)

$(z_1,z_2) = z^T_1z_2$

$=\begin{bmatrix}x_1 & y_1\\\end{bmatrix}\begin{bmatrix}x_2\\y_2\end{bmatrix}=\Bigl(x_1x_2+y_1y_2\Bigr)$

Exercise $$\PageIndex{1}$$

Prove $$(z_1,z_2)=(z_2,z_1)$$.

When $$z_1=z_2=z$$, then the inner product between $$z$$ and itself is the norm squared of $$z$$:

$||z||^2=(z,z)=x^2+y^2$

These properties of vectors seem abstract. However, as we now show, they may be used to develop a vector calculus for doing complex arithmetic.

## A Vector Calculus for Complex Arithmetic

The addition of two complex numbers $$z_1$$ and $$z_2$$ corresponds to the addition of the vectors $$z_1$$ and $$z_2$$:

$z_1+z_2⇔z_1+z_2=\begin{bmatrix}x_1+x_2\\y_1+y_2\end{bmatrix}$

The scalar multiplication of the complex number $$z_2$$ by the real number $$x_1$$ corresponds to scalar multiplication of the vector $$z_2$$ by $$x_1$$

$x_1z_2⇔x_1\begin{bmatrix}x_2\\y_2\end{bmatrix}=\begin{bmatrix}x_1x_2\\x_1y_2\end{bmatrix}$

Similarly, the multiplication of the complex number $$z_2$$ by the real number $$y_1$$ is

$y_1z_2⇔y_1\begin{bmatrix}x_2\\y_2\end{bmatrix}=\begin{bmatrix}y_1x_2\\y_1y_2\end{bmatrix}$

The complex product $$z_1z_2 = (x_1+jy_1)z_2$$ is therefore represented as

$z_1z_2↔\begin{bmatrix}x_1x_2−y_1y_2\\x_1y_2+y_1x_2\end{bmatrix}$

This representation may be written as the inner product

$z_1z_2=z_2z_1↔\begin{bmatrix}(v,z_1)\$$w,z_1)\end{bmatrix}$ where v and w are the vectors \(v=\begin{bmatrix}x_2\\−y_2\end{bmatrix}$$ and $$w=\begin{bmatrix}y_2\\x_2\end{bmatrix}$$. By defining the matrix

$\begin{bmatrix}x_2 & −y_2\\y_2 & x_2\end{bmatrix}$

we can represent the complex product $$z_1z_2$$ as a matrix-vector multiply (more on this in Linear Algebra):

$z_1z_2= z_2z_1 ↔\begin{bmatrix}x_2 & −y_2\\y_2 & x_2\end{bmatrix}\begin{bmatrix}x_1\\y_1\end{bmatrix}$

With this representation, we can represent rotation as

$ze^{jθ}= e^{jθ}z ↔\begin{bmatrix}\cosθ & −\sinθ\\\sinθ & \cosθ\end{bmatrix}\begin{bmatrix}x_1\\y_1\end{bmatrix}$

We call the matrix $$\begin{bmatrix}\cosθ & −\sinθ\\\sinθ & \cosθ\end{bmatrix}$$ a “rotation matrix.”

Exercise $$\PageIndex{2}$$

Call $$\mathrm R (θ)$$ the rotation matrix:

$\mathrm R (θ)=\begin{bmatrix}\cosθ & −\sinθ\\\sinθ & \cosθ\end{bmatrix}$

Show that $$\mathrm R (−θ)$$ rotates by $$(−θ)$$. What can you say about $$\mathrm R (−θ)w$$ when $$w=\mathrm R (θ)z$$?

Exercise $$\PageIndex{3}$$

Represent the complex conjugate of $$z$$ as

$z^∗↔\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

and find the elements $$a,b,c,$$ and $$d$$ of the matrix.

## Inner Product and Polar Representation

From the norm of a vector, we derive a formula for the magnitude of z in the polar representation $$z=re^{jθ}$$

$r=(x^2+y^2)^{1/2} = ||z|| = (z,z)^{1/2}$

If we define the coordinate vectors $$e_1=\begin{bmatrix}1\\0\end{bmatrix}$$ and $$e_2=\begin{bmatrix}0\\1\end{bmatrix}$$, then we can represent the vector $$z$$ as

$z=(z,e_1)e_1 + (z,e_2)e_2$

See Figure. From the figure it is clear that the cosine and sine of the angle $$θ$$ are

$\cosθ=\frac {(z,e1)} {||z||};\; \sinθ = \frac {(z,e_2)} {||z||}$

This gives us another representation for any vector z:

$z=||z||\cosθe_1+||z||\sinθe_2$

The inner product between two vectors $$z_1$$ and $$z_2$$ is now

$(z_1,z_2)=[(z_1,e_1)e^T_1(z_1,e_2)e^T_2]\begin{bmatrix}(z_2,e_1)e_1\$$z_2,e_2)e_2\end{bmatrix}$ $=(z_1,e_1)(z_2,e_1)+(z_1,e_2)(z_2,e_2)$ $=||z_1||cosθ_1||z_2||cosθ_2+||z_1||sinθ_1||z2||sinθ_2$ It follows that \(\cos(θ_2−θ_1)=\cosθ_2 \cos θ_1+\sinθ_1\sinθ_2$$ may be written as

$\cos(θ_2−θ_1)=\frac {(z_1,z_2)} {||z_1||\,||z_2||}$

This formula shows that the cosine of the angle between two vectors $$z_1$$ and $$z_2$$, which is, of course, the cosine of the angle of $$z_2z^∗_1$$, is the ratio of the inner product to the norms.

Exercise $$\PageIndex{4}$$

Prove the Schwarz and triangle inequalities and interpret them:

$(\mathrm {Schwarz})\; (z_1,z_2)^2≤||z_1||^2||z_2||^2$
$(\mathrm {Triangle})\; I\,||z_1−z_2||≤||z_1−z_3||+||z_2−z_3||$