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Polar Form

You should be familiar with Complex Numbers.

The complex numbers can be represented as a point (or vector) in a 2-dimensional Cartesian Coordianate system, called the Argand Diagram. The complex number \( z = x + iy \) is represented by the point \( P_z = (x, y ) \).

{{image PolarForm }}

The distance of the point to the origin \( (O) \) is known as the absolute value (or modulus), and is equal to \( r_z = \sqrt{x^2 + y^2 } \). The angle between line \( P_z O \) and the positive real axis is known as the argument (or phase) of \(z\), which we denote by \( \theta_z \). From the definition, we easily see that \( \theta_z\) is related to \( \arctan \left( \frac{y}{x} \right) \), where slight care has to be taken when the angle is obtuse, \( x=0 \) or \( y = 0 \). Typically, \( \theta_z\) is expressed in radians, and represented by the principal value in the interval \( \left( - \pi, \pi \right] \).

With these values of \( r_z\) and \( \theta_z \), we see that

\[ z = x + iy = r_z ( \cos \theta_z + i \sin \theta_z ). \]

This is known as the polar form of \(z\), which is sometimes abbreviated to \( z = r_z \text{ cis } \theta_z \).

Euler's formula: For any real number \(x\), \[ e^{ix} = \cos x + i \sin x. \]

As such, we may also express the complex number \(z\) as \( r_z e^{ i \theta_z } \). With this interpretation, multiplication and division of complex numbers follow the rules of exponentiation that we are used to. Given two complex numbers \( z_1 = r_1 \text{ cis } \theta_1 = r_1 e^{i \theta_1} \) and \( z_ 2 = r_2 \text{ cis } \theta_2 = r_2 e^{i \theta_2}\), we have

\[ z_1 z_2 = (r_1 r_2) e^{ i ( \theta_1 + \theta_2 ) } = ( r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2), \] \[ \frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i (\theta_1 - \theta_2) } = \frac{r_1}{r_2} \text{ cis } (\theta_1 - \theta_2).\]

While multiplication and division are greatly simplified, addition and subtraction of complex numbers in polar form do not follow a standard algorithm. Thus, in calculations involving complex numbers, it is important to be able to properly convert between the rectangular and polar forms.

Worked Examples

1. Show that \(z_1 z_2 = ( r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2).\)

Solution: \[\begin{array} {l l} z_1 z_2 & = r_1 \text{ cis } \theta_1 \times r_2 \text{ cis } \theta_2 \\ & = r_1 r_2 ( \cos \theta_1 + i \sin \theta_1 ) ( \cos \theta_2 + i \sin \theta_2 ) \\ & = r_1 r_2 \left( \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i ( \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2 ) \right)\\ & = r_1 r_2 \left( \cos (\theta_1 + \theta_2) + i \sin ( \theta_1 + \theta_2) \right) \\ & = (r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2) \\ \end{array} \]

The division of complex numbers in polar form can be dealt with in a similar manner, or by noticing that \( \frac{1}{z_2} = \frac{1}{r_2} \text{ cis } (- \theta_2 ) \).

 

2. Evaluate \( ( 2 + i) ( 3 + i ) \). Hence, determine \( \arctan \frac{1}{2} + \arctan \frac{1}{3} \).

Solution: Let \( z_1 = 2 +i, z_2 = 3+i, z_3 = z_1 \times z_2 \). Expanding \(z_3\), we get \[ z_3 = (2+i)(3+i) = 6 + 2i + 3i + i^2 = 5 + 5i . \]

By comparing the arguments on both sides, we get that \( \theta_1 + \theta_2 = \theta_3\), which gives us

\[ \arctan \frac{1}{2} + \arctan \frac{1}{3} = \arctan \frac{5}{5} = \frac{\pi}{4}. \]

Note by Arron Kau
3 years ago

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i want explanation for second example can anyone help me Rishabh Jain · 2 years, 9 months ago

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