How To Evaluate Complex Numbers

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To evaluate the complex number Z = (3-4i)^5, it is suggested to use DeMoivre's theorem, which states that (r[cos(θ) + i sin(θ)])^n = r^n (cos(nθ) + i sin(nθ)). However, for simplicity, repeatedly multiplying the complex number by itself is also a viable method, especially for integer exponents. The discussion emphasizes that understanding polar representation is useful but not strictly necessary for this evaluation. Ultimately, both methods can lead to the correct result, with the multiplication approach being more straightforward for those unfamiliar with polar coordinates. The conversation concludes with encouragement to choose the method that feels most comfortable.
urduworld
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Hi PFs



i want to know how to evaluate the complex number, and what are the meaning of the evaluating a complex number



Let Z be the complex number, Z = (3-4i)^5 what i have to do, just give me hint
 
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urduworld said:
Hi PFs



i want to know how to evaluate the complex number, and what are the meaning of the evaluating a complex number



Let Z be the complex number, Z = (3-4i)^5 what i have to do, just give me hint

Put the complex number into its polar representation: r.e.
 
its polar representation is Z = r[Cos(thita) + iSin(thita)]
is it is Evaluation
i want to know what actually i have to do in evaluation
Thanks for your help :)
 
urduworld said:
its polar representation is Z = r[Cos(thita) + iSin(thita)]
is it is Evaluation
i want to know what actually i have to do in evaluation
Thanks for your help :)

You need to go in the reverse direction, using x and y as your variables, not r and θ. Can you complete the following with substitutions for r and θ in terms of x and y?

r^{i\theta} = x + iy​

Alternatively, can you rewrite 3-4i into polar form, to, say, four decimal places?

Cheers -- sylas
 
i was thinking that evaluation is like polar form i don't know how to resolve this, i know nothing about
r^{i\theta} = x + iy
please tell me what i have to do,
 
To find powers you need "DeMoivre's formula":
(r[cos(\theta)+ i sin(\theta)])^n= r^n (cos(n\theta)+ i sin(n\theta)).

If DeMoivre's formula is not in your text, did you consider just sitting down and multiplying 3- 4i by itself 5 times? That would probably have been faster.
 
oOps this is like that yes Demorvies theorem is in mine book, i have to apply this, and is n will the power like 0,1,2...n
 
urduworld said:
i was thinking that evaluation is like polar form i don't know how to resolve this, i know nothing about
r^{i\theta} = x + iy
please tell me what i have to do,

I fool so feelish...

HallsofIvy is quite right; simply multiplying it is probably the easiest thing to do when you have an integer exponent.

For example:
(3-4i)(3-4i) = 9 - 24i + 16i^2 = 9 - 16 - 24i = -7 -24i​
Keep going from there. Multiply by (3-4i), and again, and once more. You don't need the polar form.

However, since I mentioned it, I'd better go ahead with the answer:
\sqrt{x^2+y^2} \times e^{\text{atan2}(y, x) i} = x + iy​
You don't need to worry about this if you take HallsofIvy's hint about repeated multiplications.

Cheers -- sylas
 

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