Expression of a Complex Number in Polar Form

By using the Trigonometric ratios \cos{\theta}=\frac{x}{r} and \sin{\theta}=\frac{y}{r} we can express any Complex number z=x+iy in Polar form. We can do this by rearranging the equations for cosine and sine by saying x=r\cos{\theta} and y=r\sin{\theta} and then state a Complex number as

z=x+iy=r\cos{\theta}+ir\sin{\theta}

with r=|z|=\sqrt{x^{2}+y^{2}} the modulus or the distance from the origin to the point z=(x,y)

and \theta=arg(z)=tan^{-1}\frac{y}{x} the argument or the directed angle to the vector representing z as measured counterclockwise from the positive x-axis.

Example

z=\sqrt{3}+i and r=|z|=\sqrt{(\sqrt{3})^{2}+(1)^{2}}=\sqrt{3+1}=\sqrt{4}=2

\theta has \cos{\theta}=\frac{\sqrt{3}}{2} and \sin{\theta}=\frac{1}{2}

both x>0 and y>0, so the angle is in Quadrant I and arg(z)=\tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}

then

z=2\cos{\frac{\pi}{6}}+2i\sin{\frac{\pi}{6}}=2(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}})

with

x=2(\frac{\sqrt{3}}{2})=\sqrt{3} and y=2(\frac{1}{2})=1

(30:60:90 triangle)

Example

z=1+\sqrt{3}i and r=\sqrt{(1)^{2}+(\sqrt{3})^{2}}=2

with \theta=arg(z)=\tan^{-1}(\frac{\sqrt{3}}{1})=\frac{\pi}{3}

then

z=2(\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}})

with

x=2(\frac{1}{2})=1 and y=2(\frac{\sqrt{3}}{2})=\sqrt{3}

(60:30:90 triangle)

Example

z=-2-2i with r=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}

both x=-2 and y=-2 are negative so the angle is located in the 3rd Quadrant and we must be careful because the tangent function has period of \pi and the cosine and sine functions have periods of 2\pi

\theta=arg(z)=\tan^{-1}(\frac{-2}{-2})=\frac{5\pi}{4} or \pi + \frac{\pi}{4}=\frac{4\pi}{4}+\frac{\pi}{4}=\frac{5\pi}{4}

then

z=2\sqrt{2}(\cos{\frac{5\pi}{4}}+i\sin{\frac{5\pi}{4}}

with

x=2\sqrt{2}(\frac{-\sqrt{2}}{2})=-2 and y=2\sqrt{2}(\frac{-\sqrt{2}}{2})=-2

(45:45 in third quadrant)

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Multiplication of the Polar Form and Angle Addition Formulas

Multiplying Complex numbers in Polar form gives insight into how the angle of the Complex number changes in an explicit way.

We know from the section on Multiplication that when we multiply Complex numbers, we multiply the components and their moduli and also add their angles, but the addition of angles doesn't immediately follow from the operation itself. The new Complex number and its modulus do, but the addition of angles needs to be worked out.

Of course, we can see through analysis that this is the case, but it is not until we see Complex numbers in Polar form that we see that it is natural to add the angles.

Here is an explicit proof that angle addition is a result of multiplication:

We need to apply the Trigonometric rules for the addition of angles

\cos{\theta_1+\theta_2}=\cos{\theta_1}\cos{\theta_2}-\sin{\theta_1}\sin{\theta_2}

and

\sin{\theta_1+\theta_2}=\cos{\theta_1}\sin{\theta_2}+\cos{\theta_2}\sin{\theta_1}

Example

If z_1=r_1(\cos{\theta_1}+i\sin{\theta_1}) and z_2=r_2(\cos{\theta_2}+i\sin{\theta_2}) then their product is

z_1z_2=r_1(\cos{\theta_1}+i\sin{\theta_1})r_2(\cos{\theta_2}+i\sin{\theta_2})=

r_1r_2[(\cos{\theta_1}\cos{\theta_2}-\sin{\theta_1}\sin{\theta_2})+i(\cos{\theta_1}\sin{\theta_2}+\cos{\theta_2}\sin{\theta_1})]=

r_1r_2[\cos{(\theta_1+\theta_2)}+i\sin{(\theta_1+\theta_2)}]

This shows that multiplying 2 Complex numbers results in the addition of the angles of the numbers.

We can immediately apply the formula to show that the product of a Complex number z with itself zz=z^{2} has a special consequence

z^{2}=r^{2}(\cos{(\theta+\theta)}+i\sin{(\theta+\theta)})=r^{2}(\cos{2\theta}+i\sin{2\theta})

Example

If z=1-\sqrt{3}i with r=\sqrt{(1)^{2}+(\sqrt{3})^{2}}=2 and \theta=\frac{5\pi}{3}\implies 2\theta=\frac{10\pi}{3} which is equivalent to \frac{4\pi}{3}

z^{2}=(2)^{2}(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}})=4(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)=-2-2\sqrt{3}i

Example

z=-3+3i with r=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{18}\implies r^{2}=18 and \theta=\frac{3\pi}{4}\implies 2\theta=\frac{3\pi}{2}

z^{2}=18(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}})=18(0-i)=-18i

Division of Polar Form and Angle Subtraction Formulas

The Complex Exponential Function

The exponential number e raised to a Complex number z is more easily handled when we convert the Complex number to Polar form

e^{z}=e^{x+iy}=e^{x}e^{iy}=e^{x}(\cos{y}+i\sin{y})

where e^{x} is the Real part and is the radius or modulus and e^{iy}=(\cos{y}+i\sin{y}) is the Imaginary part with y as the argument.

If z=0+iy then e^{iy}=\cos{y}+isin{y} becomes $e^{i\theta}=\cos{\theta}+i\sin{\theta}

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Patrick