NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Conversion and Simplification involving Polar form
Lesson

 

A conceptual view of multiplication and division

In polar form multiplication can be seen in terms of transformations. Each complex number has two independent parts that determine its location on the complex plane.

Often visualised as an arrow, the crucial elements of the complex number $z=r\left(\cos\theta+i\sin\theta\right)=r\operatorname{cis}\theta$z=r(cosθ+isinθ)=rcisθ are its modulus $r=\left|z\right|$r=|z| (depicted as the length of the arrow) and its argument $\theta$θ (depicted as the angle the arrow makes with the positive real axis and denoted as $Arg$Arg $z$z). 

In polar form if we multiply $z_1$z1 by $z_2$z2, the result is a complex number with a modulus given as the product of the moduli of $z_1$z1 and $z_2$z2 and an argument given as the sum of the arguments of $z_1$z1 and $z_2$z2

This essentially means that the result is a dilated and rotated arrow as shown schematically here:  

 

As a simple example, multiplication of the complex numbers $z_1=-2$z1=2 and $z_2=-3$z2=3, each with an argument of $\pi$π radians (or $180^{\circ}$180 ) and having moduli of $2$2 and $3$3 respectively, results in the complex number $6$6, with modulus $6$6 and argument of $2\pi$2π radians ( $360^{\circ}=0^{\circ}$360=0). 

Division therefore involves a division of the moduli and a subtraction of the arguments. So, for example, we have $\frac{6\operatorname{cis}45^{\circ}}{3\operatorname{cis}15^{\circ}}=2\operatorname{cis}30^{\circ}$6cis453cis15=2cis30.   

 

General results

We can write down general statements for the multiplication and division of two complex numbers that express this transformational idea.

So given:

$z_1=r_1\left(\cos\theta_1+i\sin\theta_1\right)=r_1\operatorname{cis}\theta_1$z1=r1(cosθ1+isinθ1)=r1cisθ1

$z_2=r_2\left(\cos\theta_2+i\sin\theta_2\right)=r_2\operatorname{cis}\theta_2$z2=r2(cosθ2+isinθ2)=r2cisθ2

we have:

$z_1\times z_2=r_1r_2\operatorname{cis}\left(\theta_1+\theta_2\right)$z1×z2=r1r2cis(θ1+θ2)

$\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}\left(\theta_1-\theta_2\right)$z1z2=r1r2cis(θ1θ2)

 
 
Example 1

Evaluate the product $z=2\operatorname{cis}\frac{\pi}{4}\times\frac{1}{2}\operatorname{cis}\left(-\pi\right)$z=2cisπ4×12cis(π)

The angle measure is in radians, but the question could have just as easily been given using degrees. The product becomes $2\times\frac{1}{2}\times\operatorname{cis}\left(\frac{\pi}{4}-\pi\right)=\operatorname{cis}\left(-\frac{3\pi}{4}\right)$2×12×cis(π4π)=cis(3π4).

Example 2

Simplify $z=\frac{12\left(\cos15^{\circ}+i\sin15^{\circ}\right)}{3\left[\cos\left(-60^{\circ}\right)-i\sin\left(-60^{\circ}\right)\right]}$z=12(cos15+isin15)3[cos(60)isin(60)].

Noting that $\cos\left(-\theta\right)=\cos\left(\theta\right)$cos(θ)=cos(θ) and $\sin\left(-\theta\right)=-\sin\left(\theta\right)$sin(θ)=sin(θ) we can immediately simplify the denominator so that:

$z=\frac{12\left(\cos15^{\circ}+i\sin15^{\circ}\right)}{3\left(\cos60^{\circ}+i\sin60^{\circ}\right)}$z=12(cos15+isin15)3(cos60+isin60)

Then we have:

$z$z $=$= $\frac{12\left(\cos15^{\circ}+i\sin15^{\circ}\right)}{3\left(\cos60^{\circ}+i\sin60^{\circ}\right)}$12(cos15+isin15)3(cos60+isin60)
  $=$= $4\left[\operatorname{cis}\left(-45^{\circ}\right)+i\sin\left(-45^{\circ}\right)\right]$4[cis(45)+isin(45)]
     

We can depict $z=4\operatorname{cis}\left(-45^{\circ}\right)$z=4cis(45) geometrically as an arrow of length $4$4 emanating from the origin into the $4$4th quadrant.  

 
 
Example 3

Explain the multiplication $\left(1+i\right)\left(1+\sqrt{3}i\right)$(1+i)(1+3i) visually.

We can approach the problem in a number ways.

We could first convert each complex factor into polar form and use the transformational idea to show the result on the complex plane. Since $1+i=\sqrt{2}\operatorname{cis}45^{\circ}$1+i=2cis45 and $1+\sqrt{3}i=2\operatorname{cis}60^{\circ}$1+3i=2cis60, the product becomes $2\sqrt{2}\operatorname{cis}105^{\circ}$22cis105 . Using radians, we could also write this as $2\sqrt{2}\operatorname{cis}\frac{7\pi}{12}$22cis7π12.

An alternative strategy is to use the fact that multiplication of any non-zero complex number by $i$i is equivalent to a simple $90^{\circ}$90 anticlockwise rotation of that complex number in the complex plane.

Thus our strategy begins by partially expanding the the given product so that:

$\left(1+i\right)\left(1+\sqrt{3}i\right)$(1+i)(1+3i) $=$= $1\left(1+\sqrt{3}i\right)+i\left(1+\sqrt{3}i\right)$1(1+3i)+i(1+3i)
     

The product can now be seen as the multiplication of $1$1 by $\left(1+\sqrt{3}i\right)$(1+3i) added to the $90^{\circ}$90 rotation of the arrow representing $\left(1+\sqrt{3}i\right)$(1+3i).

We can picture this as the sum of two arrows, and, being isomorphic to vector addition use the parallelogram law of vectors to represent the result.  

 

Worked Examples

Question 1

Find the value of $\frac{6\left(\cos220^\circ+i\sin220^\circ\right)}{3\left(\cos190^\circ+i\sin190^\circ\right)}$6(cos220°+isin220°)3(cos190°+isin190°) in rectangular form.

Question 2

Find the value of $\frac{12}{\sqrt{3}+i}$123+i in rectangular form by first converting the numerator and denominator to polar form.

Question 3

Find the value of $\frac{2\sqrt{2}+2\sqrt{6}i}{-\sqrt{6}+\sqrt{2}i}$22+26i6+2i in rectangular form by first converting the numerator and denominator to polar form.

Outcomes

M8-9

Manipulate complex numbers and present them graphically

91577

Apply the algebra of complex numbers in solving problems

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