Hcc8.11 change each to complex form and find product

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SUMMARY

The discussion focuses on converting complex numbers into different forms and finding their product using De Moivre's Theorem. The product of the complex numbers $(1+3i)$ and $(2-2i)$ is calculated as $8 + 4i$. Participants clarify the terminology, emphasizing the distinction between Cartesian form and Polar form, where Polar form is expressed as "r(cos(θ) + i sin(θ))" or "r cis(θ)". Key calculations involve determining the modulus r and the angle θ using the formulas r = √(a² + b²) and θ = tan⁻¹(b/a).

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karush
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$\tiny{hcc8.11}$
$\textsf{Find product $(1+3i)(2-2i)$}\\$

$8 + 4i$
$\textsf{Then change each to complex form and find product. with DeMoine's Theorem}$

$\textit{ok looked at an example but ??}
 
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What you're asking is not clear. Can you quote directly from your source? Do you mean "DeMoivre's theorem"?
 
basically yes
but this got answered

it was on a hand out which was hard to read with little information.
 
You probably should not use the phrase "complex form" here. These are complex number which are typically written in one of two forms, "Cartesian form", which is what you have, and "Polar form", "r(cos(\theta)+ i sin(\theta))" or (an engineering notation) "r cis(\theta)". For "a+ bi", r= \sqrt{a^2+ b^2} and \theta= tan^{-1}(b/a) (as long as a is not 0. If a= 0 \theta= \pi/2 (if b> 0) or \theta= -\pi/2 if b< 0). if a=b= 0, r= 0 and \theta can be anything.
 
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