Multiplying Complex Numbers In Trigonometric Form

Ted_Grendy

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Joined
Nov 11, 2018
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36
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

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Nov 24, 2012
Messages
1,816
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?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,319
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
Joined
Nov 12, 2017
Messages
4,110
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.
 
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