3EPolarArithmetic


 * Arithmetic in Polar Form **

Addition and Subtraction
 * Best done in Cartesian Form

Multiplication

math . \qquad \text {For } z_1 =r_1 \, \text {cis} \big( \theta_1 \big) \; \text { and } \; z_2 = r_2 \, \text{cis} \big( \theta_2 \big) math

math . \qquad z_1 \times z_2 = \left( r_1r_2 \right) \, \text {cis} \left( \theta_1 + \theta_2 \right) math


 * Example 1 **

math \text{Simplify } \quad 2 \, \text{cis} \left( \dfrac{\pi}{4} \right) \times 3 \, \text{cis} \left( \dfrac{\pi}{3} \right) math


 * Solution:**

math . \qquad 2 \, \text{cis} \left( \dfrac{\pi}{4} \right) \times 3 \, \text{cis} \left( \dfrac{\pi}{3} \right) = 6 \, \text{cis} \left( \dfrac{7\pi}{12} \right) math

Recall

An alternate way to write a complex number in polar form is using index form. math . \qquad z=r \, \text{cis} \big( \theta \big) = r \, e^{\theta i} math

Multiplying numbers in index form involves adding the indices. math . \qquad z_1 \times z_2 = (r_1 r_2) e^{\theta_1 i + \theta_2 i} math

So we see that the polar form multiplication rule of multiplying the r values and adding the angles comes from basic index laws.

Division

math . \qquad \text {For } z_1 =r_1 \, \text {cis} \big( \theta_1 \big) \; \text { and } \; z_2 = r_2 \, \text{cis} \big( \theta_2 \big) math

math . \qquad z_1 \div z_2 = \left( \dfrac {r_1}{r_2} \right) \, \text {cis} \left( \theta_1 - \theta_2 \right) math


 * Example 2 **

math \text{Simplify } \quad 2 \, \text{cis} \left( \dfrac{\pi}{4} \right) \div 3 \, \text{cis} \left( \dfrac{\pi}{3} \right) math


 * Solution:**

math . \qquad 2 \, \text{cis} \left( \dfrac{\pi}{4} \right) \div 3 \, \text{cis} \left( \dfrac{\pi}{3} \right) = \dfrac{2}{3} \, \text{cis} \left( \dfrac{-\pi}{12} \right) math

NOTE: Be careful when adding or subtracting angles to stay within the Principal Domain (– p, p ]. Add or subtract 2 p to return an angle to within the restrictions.

Congugate of z

math . \qquad \text {For } z = r \, \text{cis} \big( \theta \big) math

The complex conjugate of z is: math . \qquad \bar{z} = r \, \text{cis} \left( -\theta \right) math

De Moivre's Theorem

De Moivre's Theorem gives us z n

math . \qquad \text {For } z = r \, \text{cis} \big( \theta \big) math

De Moivre's Theorem is math . \qquad z^n = r^n \, \text{cis} (n\theta) \quad n \in Q \qquad \{ \text {Rational numbers} \} math


 * Example 1 **

math \\ \text{Given } z = 3 \, \text{cis} \left( \dfrac{\pi}{3} \right) \\. \\ \text{Find } z^5 math

__**Solution**__

math . \qquad z^5 = 3^5 \, \text{cis} \left( \dfrac{5\pi}{3} \right) math

but the angle is outside the Principal Domain, so subtract 2 p

math . \qquad z^5 = 243 \, \text{cis} \left( -\dfrac{\pi}{3} \right) math


 * Example 2 **

math \\ \text{Given } z = 2 + 2i \\. \\ \text{Find } z^4 math

__**Solution:**__


 * We could solve this using the Binomial Theorem to expand the quartic.
 * A much easier way is to convert to Polar Form and then use De Moivre's Theorem.

math \\ . \qquad r = \sqrt{2^2+2^2} \\. \\ . \qquad \quad= \sqrt{8} \\. \\ . \qquad \theta = \text{Tan}^{-1} \left( \dfrac{2}{2} \right) \\. \\ . \qquad \quad = \text{Tan}^{-1} \big( 1 \big) \\. \\ . \qquad \quad = \dfrac{\pi}{4} math

Therefore: math . \qquad z = \sqrt{8} \, \text{cis} \left( \dfrac{\pi}{4} \right) math

Now use DeMoivre's Theorem math \\ . \qquad z^4 = \left( \sqrt{8} \right)^4 \text{cis} \left( \dfrac{4\pi}{4} \right) \\. \\ . \qquad \quad = 64 \, \text{cis} (\pi) \\. \\ . \qquad \quad = 64 \big( -1 + 0i \big) \\. \\ . \qquad \quad = -64 math

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