Hi everyone

got confused again.

3+4i^(3/2)=

THis is what i did

z^(3/2) = 5^(3/2) = ( cos(3(2pik+0.9273))/2 + isin3((2pik+0.9273))/2 )

Please advise, a bit confuse with the the ^3/2 root. should we times the whole (2pik + titah)/2 equation with 3/2 ?

For k=0 i get

=11.17(cos1.39095 + isin1.39095)

k = 1

=11.17(cos10.8157 + isin10.8157)

k=2

=11.17(cos20.24 = isin 20.24)

Pls comment, im revising for my final.thank you & regards. so sorry , i got confused with the ^3/2 substitution.

thank you & regards