What are the values of x and y for the complex number z=\sqrt{3+4i}?

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To find the values of x and y for the complex number z = √(3 + 4i), start by squaring both sides of the equation, leading to x + yi = √(3 + 4i). Equate the real and imaginary parts of both sides to solve for x and y. There are two square roots for this complex number, indicating two possible solutions. Alternatively, the problem can be approached using polar form and de Moivre's theorem, though the direct method is simpler. Ultimately, the discussion emphasizes the importance of equating parts and recognizing the dual solutions in complex numbers.
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z=x+yi determine the values of x and y such that z=\sqrt{3+4i}
I'm not even sure where to start with this one, so any help would be greatly appreciated
 
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So basically the question you are given is find x and y where
x+yi=\sqrt{3+4i}.
Start off by squaring both sides, then go about solving for x and y. When solving variables with complex numbers, you usually equate the real and imaginary parts of both sides. In other words, if you have x+iy=u+iv, then x=u and y=v. This should get you going. Let us know if you are still stuck.
 
In other words, find the square root of 3+4i. Of course, there are two square roots - the problem said "values"- so n!kofeyn's equations will have two solutions. You could also do this problem by converting to "polar" form and applying deMoivre's formula. That was my first thought but n!kofeyn's idea is simpler and more straightforward.
 
Awesome. I've got it now. Thanks for your help.
 
No problem.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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