Thread: Complex cube root

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

Originally Posted by anderson

Hi everyone

got confused again.

3+4i^(3/2) =

This is not a "cube root"

THis is what i did

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

You mean 5^(3/2) times that not "=".

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

You should multiply by 3. That "/2" is the denominator, isn't it? Multiply by 3/2 which is the same, of course, as "multiply by 3 and divide by 2".

For k=0 i get
=11.17(cos1.39095 + isin1.39095)[/quote]
, to two decimal places, is 11.18, not 11.17.

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

Those numbers are right.