# Complex plane#

Mathematics Methods 2

From the definition of complex numbers it is clear that there is a natural correspondence between a set of complex numbers $$\mathbb{C}$$ and points in a plane: to every complex number $$z = x +iy$$ we can uniquely assign a point $$P(x, y)$$ in the $$\mathbb{R}^2$$ plane. We call a plane in which every point has a corresponding complex number assigned to it a complex plane (or Argand plane). However, for simplicity we would normally simply say “in a point $$z$$” when we mean “in a point (of a complex plane) assigned a complex number $$z$$.”

z = complex(3, 2)

fig = plt.figure()

plt.plot(z.real, z.imag, 'o', zorder=10)
plt.plot([3, 3], [0, 2], '--k', alpha=0.8)
plt.plot([0, 3], [2, 2], '--k', alpha=0.8)

ax.text(2.7, -0.25, "$Re(z)$", fontsize=12)
ax.text(-0.7, 1.95, "$Im(z)$", fontsize=12)

ax.axis('equal')
ax.set_xlim(-2, 4)
ax.set_ylim(-2, 4)
ax.axis('off')

plt.show() Having represented complex numbers geometrically, let us revisit the terms we introduced in the first notebook.

• Re $$z$$ is equal to the abscissa and Im $$z$$ to the ordinate of a point $$P$$. We therefore call x-axis the real axis and y-axis the imaginary axis.

• The modulus of a complex number $$|z|$$ is equal to the distance between the corresponding point $$P$$ and the origin of the complex plane. In general, the distance between points $$P_1$$ and $$P_2$$ assigned to numbers $$z_1$$ and $$z_2$$ is equal to the modulus of the difference of those two points:

$d(T_1, T_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = | z_1 - z_2 |$
• If a number $$z$$ is assigned a point $$P(x, y)$$, then the complex conjugate $$z^*$$ is assigned a point $$P^*(x, -y)$$, which is just a reflection of $$P$$ with respect to the real axis.

• Every point $$P(x, y)$$ has a corresponding vector $$\vec{OP}$$ from the origin to $$P$$. The sum $$z_1 + z_2$$ therefore represents vector addition $$\overrightarrow{OP_1} + \overrightarrow{OP_2}$$.

Because of the vector analogy, we can also say that complex numbers satisfy the triangle inequalities:

$| z_1| - |z_2| \leq |z_1 + z_2| \leq |z_1| + |z_2|$

# Trigonometric form#

We can also observe the complex plane in polar coordinates $$(r, \phi)$$ which are related to the Cartesian coordinates $$(x, y)$$ through:

$x = r \cos \varphi, \quad y = r \sin \varphi$

This leads to the trigonometric form of a complex number $$z$$:

$z = r(\cos \varphi + i \sin \varphi),$

where $$r$$ is the modulus (or magnitude), equal to the absolute value of the complex number, a direct consequence of Pythagora’s theorem:

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

and the polar angle $$\varphi$$ is defined (to a multiply $$2\pi$$) by:

$\tan (\varphi + 2n \pi) = \frac{y}{x}, \quad n \in \mathbb{Z}, x \neq 0.$

The angle $$\varphi$$ is called the argument (or phase) of a complex number $$z$$ and we write $$\text{Arg}(z) = \varphi$$. A special case is the complex number $$z=0$$ for which $$r = 0$$ and its argument is not defined.

Finding $$\varphi$$ is delicate because $$\tan$$ is a multivalued function. To avoid ambiguity, the simplest choice is $$n = 0$$ so that the interval is of length $$2\pi$$ and $$- \pi < \text{arg}(z) \leq \pi$$. The value of $$\text{Arg}(z)$$ with $$n=0$$ is called the principal value of the argument. With this:

$\text{arg}(1) = 0, \quad \text{arg}(i) = \frac{\pi}{2}, \quad \text{arg}(-1) = \pi, \quad \text{arg}(-i) = -\frac{\pi}{2}, \quad \text{etc.}$

THe relationship between $$\text{Arg}(z)$$ and $$\text{arg}(z)$$ is therefore:

$\text{Arg}(z) = \text{arg}(z) + 2n \pi, \quad n \in \mathbb{Z}.$

Multiplying two complex numbers results in their absolute values being multiplied and the arguments being added:

\begin{split} \begin{align} z_1 \cdot z_2 & = r_1(\cos \varphi_1 + i \sin \varphi_1) \cdot r_2(\cos \varphi_2 + i \sin \varphi_2) \\ & = r_1 r_2 (\cos (\varphi_1 + \varphi_2) + i \sin(\varphi_1 + \varphi_2)) \end{align} \end{split}

For the multiplicative inverse we have

\begin{split} \begin{align} z^{-1} & = \frac{1}{r(\cos \varphi + i \sin \varphi)} \cdot \frac{cos \varphi - i \sin \varphi}{cos \varphi - i \sin \varphi} = \frac{\cos \varphi - i \sin \varphi}{r( \cos^2 \varphi + \sin^2 \varphi)} \\ & = \frac{1}{r} (\cos \varphi - i \sin \varphi) \end{align} \end{split}

Properties of $$|z|$$ and arg($$z$$)

Let us summarise our findings in the form of the following equalities:

$| z_1 \cdot z_2 | = |z_1| \cdot |z_2| = r_1 r_2, \quad \text{arg}(z_1 \cdot z_2) = \text{arg}(z_1) + \text{arg}(z_2),$
$|z^{-1}| = |z|^{-1} = r^{-1}, \quad \text{arg}(z^{-1}) = -\text{arg}(z)$

where $$z_1, z_2 \neq 0$$. By induction we get:

$|z^n| = |z|^n, \quad \text{arg}(z^n) = n\text{arg}(z), \quad \forall n \in \mathbb{Z}.$

# Polar form#

The property $$\text{arg}(z_1 \cdot z_2) = \text{arg}(z_1) + \text{arg}(z_2)$$ might remind us of logarithms, where $$\log (a \cdot b) = \log a + \log b$$. This is not a coincidence!

The exponential function, which we will look at in the next chapter, allows us to write complex numbers in a polar form:

$z = r e^{i \varphi}$

where we have used the Euler’s formula:

$e^{i \varphi} = \cos(\varphi) + i \sin (\varphi).$

In this representation, certain operations become much easier. For example,

$z_1 \cdot z_2 = r_1 r_2 e^{i(\varphi_1 + \varphi_2)}, \quad z^n = r^n e^{in \varphi}$

for all powers $$n \in \mathbb{Z}$$. If we write this using a complex number with unit modulus we recover de Moivre’s formula:

$(\cos \varphi + i \sin \varphi)^n = (e^{i \varphi})^n = e^{in \varphi} = \cos (n \varphi) + i \sin (n \varphi).$

Tip: Trigonometric identities

Trigonometric identities are also much easier to be recovered this way. For example, let us think about angle addition $$\theta + \varphi$$.

$e^{i(\theta + \varphi)} = e^{i \theta} e^{i \varphi}$

We Apply Euler’s formula to each complex number.

\begin{split}\begin{align} \cos (\theta + \varphi) + i \sin(\theta + \varphi) & = (\cos \theta + i \sin \theta)(\cos \varphi + i \sin \varphi) \\ & = \cos \theta \cos \varphi + i \cos \theta \sin \varphi + i \sin \theta \cos \varphi + i^2 \sin \theta \sin \varphi \\ & = (\cos \theta \cos \varphi - \sin \theta \sin \varphi) + i(\cos \theta \sin \varphi + \sin \theta \cos \varphi) \end{align} \end{split}

Equate real and imaginary parts on both sides:

$\begin{split}\cos (\theta + \varphi) = \cos \theta \cos \varphi - \sin \theta \sin \varphi \\ \sin (\theta + \varphi ) = \cos \theta \sin \varphi + \sin \theta \cos \varphi \end{split}$

The reader is encouraged to try to recover other trigonometric identities.