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 z = x + iy is represented by the point Pz=(x,y) P_z = (x, y ) .

{{image PolarForm }}

The distance of the point to the origin (O) (O) is known as the absolute value (or modulus), and is equal to rz=x2+y2 r_z = \sqrt{x^2 + y^2 } . The angle between line PzO P_z O and the positive real axis is known as the argument (or phase) of zz, which we denote by θz \theta_z . From the definition, we easily see that θz \theta_z is related to arctan(yx) \arctan \left( \frac{y}{x} \right) , where slight care has to be taken when the angle is obtuse, x=0 x=0 or y=0 y = 0 . Typically, θz \theta_z is expressed in radians, and represented by the principal value in the interval (π,π] \left( - \pi, \pi \right] .

With these values of rz r_z and θz \theta_z , we see that

z=x+iy=rz(cosθz+isinθz). z = x + iy = r_z ( \cos \theta_z + i \sin \theta_z ).

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

Euler's formula: For any real number xx, eix=cosx+isinx. e^{ix} = \cos x + i \sin x.

As such, we may also express the complex number zz as rzeiθz 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 z1=r1 cis θ1=r1eiθ1 z_1 = r_1 \text{ cis } \theta_1 = r_1 e^{i \theta_1} and z2=r2 cis θ2=r2eiθ2 z_ 2 = r_2 \text{ cis } \theta_2 = r_2 e^{i \theta_2}, we have

z1z2=(r1r2)ei(θ1+θ2)=(r1r2) cis (θ1+θ2), z_1 z_2 = (r_1 r_2) e^{ i ( \theta_1 + \theta_2 ) } = ( r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2), z1z2=r1r2ei(θ1θ2)=r1r2 cis (θ1θ2). \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 z1z2=(r1r2) cis (θ1+θ2).z_1 z_2 = ( r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2).

Solution: z1z2=r1 cis θ1×r2 cis θ2=r1r2(cosθ1+isinθ1)(cosθ2+isinθ2)=r1r2(cosθ1cosθ2sinθ1sinθ2+i(sinθ1cosθ2+cosθ1sinθ2))=r1r2(cos(θ1+θ2)+isin(θ1+θ2))=(r1r2) cis (θ1+θ2)\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 1z2=1r2 cis (θ2) \frac{1}{z_2} = \frac{1}{r_2} \text{ cis } (- \theta_2 ) .


2. Evaluate (2+i)(3+i) ( 2 + i) ( 3 + i ) . Hence, determine arctan12+arctan13 \arctan \frac{1}{2} + \arctan \frac{1}{3} .

Solution: Let z1=2+i,z2=3+i,z3=z1×z2 z_1 = 2 +i, z_2 = 3+i, z_3 = z_1 \times z_2 . Expanding z3z_3, we get z3=(2+i)(3+i)=6+2i+3i+i2=5+5i. z_3 = (2+i)(3+i) = 6 + 2i + 3i + i^2 = 5 + 5i .

By comparing the arguments on both sides, we get that θ1+θ2=θ3 \theta_1 + \theta_2 = \theta_3, which gives us

arctan12+arctan13=arctan55=π4. \arctan \frac{1}{2} + \arctan \frac{1}{3} = \arctan \frac{5}{5} = \frac{\pi}{4}.

Note by Arron Kau
7 years, 5 months ago

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i want explanation for second example can anyone help me

Rishabh Jain - 7 years, 3 months ago

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