Answer:
[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]
Step-by-step explanation:
The complex number given is
[tex]z = (\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2}[/tex]
Now, remember that the DeMoivre's theorem states that
[tex]( \cos(x) + i\sin(x) )^n = \cos(nx) + i\sin(nx)[/tex]
Then for this case we have that
[tex](\, \cos(\frac{\pi}{16}) + i\sin(\frac{\pi}{16}) \,)^{1/2} = \cos(\frac{\pi}{32}) + i\sin(\frac{\pi}{32}) = 0.99 + i0.09[/tex]