Finding Solutions for Complex Numbers: z^2 = 1+2i

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Homework Statement



Determine all solutions of z^2 = 1+2i in the form z=a+bi, where a and b are real numbers.For this question numerical evaluation is not required. I just don't know how to start.:mad:
any clue?
thanks!
 
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Reformulate 1+2i in the exponential form
 
static said:
Determine all solutions of z^2 = 1+2i in the form z=a+bi, where a and b are real numbers.

For this question numerical evaluation is not required. I just don't know how to start.:mad:
It seems that even if you don't have a good idea how to arrive at an answer, you have an obvious starting path: plug the answer form "z=a+bi" into the equation you're trying to solve. It may, or it may not, lead to something that works, but at least it's something to try.
 
Yep, your solution is z=a+bi, just plug it in the equation, and solve the equation. You will come up with two equations (which are actually Vieta's formulas), and you can create quadratic equation and solve for a and b.

Regards.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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