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A: Complex Roots with improper fraction

Arnaldo$$z=(-16i)^{5/4}=32(-i)^{5/4}$$ 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...

 
user163862
user163862
What happens to the -16?
 
Arnaldo
Arnaldo
the result will be multplied by $32$ like you already did.
 
user163862
user163862
I completely don't understand this. So when k=0 my answer would be $32Cis(3\pi/8)$?
 
Arnaldo
Arnaldo
@user163862: yes it is!
 
user163862
user163862
online calcs show $=-12.24-29.564i$
 
Arnaldo
Arnaldo
8:24 PM
that is the algebric form. You have to change for $cis$ form.
 
user163862
user163862
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
 
Arnaldo
Arnaldo
Actually you have $4$ solutions because $k \in \{0,1,2,3\}$.
 
user163862
user163862
do you have a moment to discuss this?
 
Arnaldo
Arnaldo
sure
 
user163862
user163862
oh thank you so much
 
Arnaldo
Arnaldo
8:25 PM
what you don't get?
 
user163862
user163862
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
 
Arnaldo
Arnaldo
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
 
user163862
user163862
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?
 
Arnaldo
Arnaldo
you can
and you will find the same solution that I did
 
user163862
user163862
I would think I could. But when I do that my first answer is 32 Cis15pi/8
which is NOT your answer
 
Arnaldo
Arnaldo
8:31 PM
see that I also find that solution
but for me is when k=3
 
user163862
user163862
ohhhhhhhhh
 
Arnaldo
Arnaldo
we both found the same solutions
but for different value of k
for example
 
user163862
user163862
so when you do this problem in one of the online calculators or with Wolfram, they list the "principal" solution as -12-29i
 
Arnaldo
Arnaldo
if you take k=1 wht you find?
 
user163862
user163862
how do they determine WHICH value is the principal?
it's actually -12.something
-12.2458698 - 29.564145 i
 
Arnaldo
Arnaldo
8:34 PM
it depends how it start for k
 
user163862
user163862
is what is listed as the principal root which I thought would be k=0
 
Arnaldo
Arnaldo
usually the principal is the complex with the smallest arg
 
user163862
user163862
ok, so just one more question
 
Arnaldo
Arnaldo
I would say it is cis(3 pi/8)
 
user163862
user163862
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
 
Arnaldo
Arnaldo
8:38 PM
I agree. Just be carefull with online calculator sometimes they have some bugs. The better way is talk to your teacher.
 
user163862
user163862
so is one of your answsers -12.2458698 - 29.564145 i
 
Arnaldo
Arnaldo
for sure it is
but it doesn't have the smallest argument
 
user163862
user163862
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?
 
Arnaldo
Arnaldo
yes
sure
 
user163862
user163862
Thank you SO very much for your help. I have to go to class now. Thank you again and Happy New Year
 
Arnaldo
Arnaldo
8:41 PM
you are very welcome
don't forget upvote the guys
and choose an answer that you liked
 

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