Simplifying and then expressing complex numbers in cartesian form

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SUMMARY

The discussion centers on simplifying the expression (2 CIS (π/6))*(3 CIS (π/12)) and converting it into Cartesian form. The correct simplification yields 6 CIS (π/4) using the properties of the CIS notation, which stands for cos(θ) + i*sin(θ). Participants clarify that CIS is an engineering notation equivalent to e^(iθ) and emphasize the application of De Moivre's theorem in this context. The final Cartesian form is derived as 6 cos(π/4) + 6i sin(π/4).

PREREQUISITES
  • Understanding of complex numbers and their representations
  • Familiarity with polar and Cartesian forms of complex numbers
  • Knowledge of De Moivre's theorem
  • Basic trigonometric functions (sine and cosine)
NEXT STEPS
  • Study the application of De Moivre's theorem in complex number operations
  • Learn about converting polar coordinates to Cartesian coordinates
  • Explore the properties of the CIS notation and its mathematical implications
  • Practice problems involving complex number multiplication and conversion
USEFUL FOR

Students studying complex numbers, mathematics educators, and anyone interested in mastering the conversion between polar and Cartesian forms of complex numbers.

Stripe
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Homework Statement


(2 CIS (pi/6))*(3 CIS (pi/12))

Homework Equations



Also what is CIS? I believe it's Cos+i*sin but how do you use it?


The Attempt at a Solution


i simplified it to

6 CIS (pi/12)

How do i turn it into cartesian?
 
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Hi Stripe! :smile:

(have a pi: π :wink:)

I've never seen "CIS" before, but I'll guess you're right, and that it's cos + i*sin.

Now use De Moivre's theorem … cosθ + isinθ = e :wink:

And Cartesian form simply means in the form x + iy (as opposed to polar form, which is in the form re )

(and no, it's not 6 CIS (π/12))
 
Last edited:
"[itex]Cis(\theta)[/itex]" is engineering notation for "[itex]cos(\theta)+ i sin(\theta)[/itex]" which mathematicians tend to write as [itex]e^{i\theta}[/itex].

The important thing about that notation is that [itex](Cis(\theta)*Cis(\phi)= Cis(\theta+ \phi)[/itex].

So you have correctly deduced that [itex](2Cis(\pi/6))(3Cis(\pi/12)= 6 Cis(3\pi/12)= 6 Cis(\pi/4)[/itex]

Now, just use the definition: [itex]6 Cis(\pi/4)= 6 cos(\pi/4)+ 6i sin(\pi/4)[/itex].
 

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