8:24 PM
0

$$z=(-16i)^{5/4}=32(-i)^{5/4}$$ Use $$-i=cis\left(\frac{3\pi}{2}+2k\pi\right)$$ Then $$(-i)^{5/4}=(-i)^{1/4}=cis\left(\frac{3\pi}{8}+\frac{1}{2}k\pi\right)$$ choose $k \in \{0,1,2,3\}$. So the solutions will be: k=0 \rightarrow z_0=32\cdot cis\left(\frac{3\pi}{8}\right)\\ k=1 \rightarro...

What happens to the -16?

the result will be multplied by $32$ like you already did.

I completely don't understand this. So when k=0 my answer would be $32Cis(3\pi/8)$?

@user163862: yes it is!

online calcs show $=-12.24-29.564i$

8:24 PM
that is the algebric form. You have to change for $cis$ form.

although I have an A in this class, this problem is completely not understandable to me in any way whatsoever.
$32 cis (3\pi/8)$ doesn't give -12.24 as the real part

Actually you have $4$ solutions because $k \in \{0,1,2,3\}$.

do you have a moment to discuss this?

sure

oh thank you so much

8:25 PM
what you don't get?

so when I do the calcs I get for k=0 16Cis15pi/8)
not 3pi /8
cause I need to multiply the 2pi/3 by 5/4
sorry need to multiply the 3pi/2 by 5/4

but first you have to sove (-i)^5
which is -i
after that you solve (-i)^{1/4}
and then you can choose k and find the proprer solution
*proper

so it's NOT possible to leave the root in improper form 5/4?
so i need to work this problem as 32(-i)^1/4?

you can
and you will find the same solution that I did

I would think I could. But when I do that my first answer is 32 Cis15pi/8

8:31 PM
see that I also find that solution
but for me is when k=3

ohhhhhhhhh

we both found the same solutions
but for different value of k
for example

so when you do this problem in one of the online calculators or with Wolfram, they list the "principal" solution as -12-29i

if you take k=1 wht you find?

how do they determine WHICH value is the principal?
it's actually -12.something
-12.2458698 - 29.564145 i

8:34 PM
it depends how it start for k

is what is listed as the principal root which I thought would be k=0

usually the principal is the complex with the smallest arg

ok, so just one more question

I would say it is cis(3 pi/8)

yes, that
that is what i thought....but cis 3pi/8 doesn't yield a negative real part nor a neg imag part
Wolfram and all the online calcs show -12.2458698 - 29.564145 i as the prin answer

8:38 PM
I agree. Just be carefull with online calculator sometimes they have some bugs. The better way is talk to your teacher.

so is one of your answsers -12.2458698 - 29.564145 i

for sure it is
but it doesn't have the smallest argument

ok, now I get this. I would not have thought to change (-16i)^(5/4) to be 32(-i)^(1/4)
but it should work without doing that, correct?

yes
sure

Thank you SO very much for your help. I have to go to class now. Thank you again and Happy New Year

8:41 PM
you are very welcome
don't forget upvote the guys
and choose an answer that you liked