How do you write #7sqrt(z^5)# as an exponential expression?

1 Answer
Jan 14, 2016

If #z# is a Complex number (as the choice of #z# rather than #x# might imply), then this can safely be written as #7(z^5)^(1/2)#, but this may or may not be equal to #7z^(5/2)#

Explanation:

Suppose #z = cos((2pi)/5) + i sin((2pi)/5)#

Then by De Moivre's formula we find:

#z^5 = cos(2pi)+i sin(2pi) = 1#

So #sqrt(z^5) = sqrt(1) = 1#

However:

#z^(5/2) = cos(pi) + i sin(pi) = -1#

So: #7(z^5)^(1/2) = 7 != -7 = 7z^(5/2)#