# Multiplying Complex Numbers In Trigonometric Form

#### Ted_Grendy

##### New member
Hi all

I was wondering if it was possible to multiple 2 complex numbers in trigonometric form.

For example if I had the following complex numbers in expressed in polar form.

z1 = 6 < 35
z2 = 2 < 25

I convert these complex numbers into trigonometric form and I get:-

z1 = 6 (cos(35) + sin(35)i)
z2 = 2 (cos(25) + sin(25)i)

My question is can I multiply the two complex trigonmetric numbers and a valid result?

I can multiply them in polar form which gives me 12 < 60 but I'm struggling with the trigonometric form.

Thank you.

#### MarkFL

##### Super Moderator
Staff member
If you multiply in trigonometric form, you get:

$$\displaystyle z_1z_2=12\left(\cos\left(35^{\circ}\right)\cos\left(25^{\circ}\right)-\sin\left(35^{\circ}\right)\sin\left(25^{\circ}\right)+i\left(\sin\left(35^{\circ}\right)\cos\left(25^{\circ}\right)+\sin\left(25^{\circ}\right)\cos\left(35^{\circ}\right)\right)\right)$$

Note that the negative sign comes from $$i^2=-1$$. If you are with me up to this point, can you now apply angle sum/difference identities to get the desired form?

• topsquark

#### pka

##### Elite Member
I was wondering if it was possible to multiple 2 complex numbers in trigonometric form.
For example if I had the following complex numbers in expressed in polar form.
Noting that $$\displaystyle \sin(\alpha+beta)=\sin(\alpha)\cos (\alpha)+\sin(\beta)\cos(\beta) ~\&~\cos(\alpha+beta)=\cos(\alpha)\cos (\beta)-\sin(\alpha)\sin(\beta)$$
\displaystyle \begin{align*}z_1z_2&=12\left(\cos\left(35^{\circ}\right)\cos\left(25^{\circ}\right)-\sin\left(35^{\circ}\right)\sin\left(25^{\circ}\right)+i\left(\sin\left(35^{\circ}\right)\cos\left(25^{\circ}\right)+\sin\left(25^{\circ}\right)\cos\left(35^{\circ}\right)\right)\right) \\&=\cos(60^o)+i\sin(60^o) \end{align*}

For ease of typing here is a shortcut: $$\displaystyle \cos(\alpha)+i\sin(\alpha)=\bf{cis(\alpha)}$$
If $$\displaystyle z_1=Tcis(\alpha)~\&~z_2=Scis(\beta)$$ then their product is $$\displaystyle z_1\cdot z_2=(TS)cis(\alpha+\beta)$$

#### Dr.Peterson

##### Elite Member
For example if I had the following complex numbers in expressed in polar form.

z1 = 6 < 35
z2 = 2 < 25

I convert these complex numbers into trigonometric form and I get:-

z1 = 6 (cos(35) + sin(35)i)
z2 = 2 (cos(25) + sin(25)i)

My question is can I multiply the two complex trigonmetric numbers and a valid result?
You are distinguishing "trigonometric form" from "polar form"; but they are really just two ways to write the same thing. On this page, both $$\displaystyle r(\cos\theta + j\sin\theta)$$ and $$\displaystyle r\cis\theta$$ and $$\displaystyle r\angle\theta$$ are identified as "polar form", and all mean the same thing.

To multiply two such numbers, you multiply the absolute values or moduli (r), and add the angles($$\displaystyle \theta$$). So the product of $$\displaystyle 6 (\cos(35^{\circ}) + \sin(35^{\circ})i)$$ and $$\displaystyle 2 (\cos(25^{\circ}) + \sin(25^{\circ})i)$$ is $$\displaystyle (2\cdot3) (\cos(25 ^{\circ}+35^{\circ}) + \sin(25^{\circ}+35^{\circ})i)= 6 (\cos(60^{\circ}) + \sin(60^{\circ})i)$$.

Others have simply shown why this is true, which applies to any of these forms.

• topsquark