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

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Homework Help Overview

The discussion revolves around finding the values of x and y for the complex number z, defined as z = √(3 + 4i). Participants are exploring the nature of complex numbers and their square roots.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss squaring both sides of the equation to solve for x and y, and equating real and imaginary parts. There is mention of the existence of two square roots for the complex number, and alternative methods such as converting to polar form are suggested.

Discussion Status

The discussion has seen some productive guidance, with participants offering different approaches to tackle the problem. There is acknowledgment of the complexity involved in finding the square roots of complex numbers.

Contextual Notes

Some participants express uncertainty about where to start, indicating a need for clarification on the problem setup and the properties of complex numbers.

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[tex]z=x+yi[/tex] determine the values of x and y such that [tex]z=\sqrt{3+4i}[/tex]
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
[tex]x+yi=\sqrt{3+4i}[/tex].
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.
 

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