When you wirte ##z=-1+0i## aren't you saying this is a pure real number once the imaginary part is zero? If so, is it right to say that the equation has a real solution. This is very confusing to me.
But if you graph the function ##y(x)=1^{x}## and look at the point ##x=0.5## you will get y=1. In the other hand, if you graph ##(-1)^{x}## you will get ##y=i##. So how can you affirm that ##1^{1/2}+1=-1+1=0##??
Thank you for the answer and the greeting! Let me put it differently:
I have the equation:
##x^{0.5}## +1 = 0
So I assume that ##x=e^{iy}## and then I get that ##e^{iy/2}+1=0##
From the Euler's identity I have a solution in the form ##y=2\pi##
So ##x=e^{2\pi i}## which gives ##x=1## and...
I got myselft wondering if there is any solution to the equation x^(0.5)+1=0. I know that for a real x there is not but, when I assume x is any imaginary number in the form e^(ix/2) and then solve the equation e^(ix/2)=-1 the result is x=2(2 pi n + pi) for integer values of n. If one then takes...