
#1
Sep1611, 05:46 AM

P: 166

In general, how many square roots does a complex number have?




#2
Sep1611, 07:01 AM

Sci Advisor
HW Helper
P: 4,301

A square root u of z is the solution to the polynomial equation
u^{2} = z (where z must be considered as a fixed number). It is a general theorem that this seconddegree polynomial has two complex roots. In fact, you can write them down explicitly: if [itex]z = r e^{i\phi}[/itex] then [tex]u_1 = \sqrt{r} e^{i\phi / 2} = \sqrt{r} \left( \cos \frac{\phi}{2} + i \sin \frac{\phi}{2} \right)[/tex] and [tex]u_2 = u_1[/tex] both square to z. 



#3
Sep1611, 07:03 AM

P: 2,068

Two  try looking up the Fundamental Theorem of Algebra.




#4
Sep1611, 07:51 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Square roots of complex numbers
In fact, it is easy to show that any nonzero complex number has precisely n distinct nth roots:
Let [itex]z= re^{i\theta}[/itex] with r> 0. Then the nth roots of z are given by [itex]r^{1/n}e^{i(\theta+ 2k\pi)/n}[/itex] where [itex]r^{1/n}[/itex] is the positive real nth root of the positive real number r and k is a nonnegative integer. For k= 0 to n1, those are distinct because [itex]0\le 2k\pi/n< 2\pi[/itex] but when k= n, [itex]2k\pi/n= 2n\pi/n= 2\pi[/itex] and [itex]e^{i(\theta+ 2\pi)}= e^{i\theta}[/itex]. 


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