# Precalc DeMoivre's Theorem to find powers of Complex numbers

"Use Demoivre's Theorem to find the indicated power of the complex number. Write the result in standard form."

:

2(squareroot of (3) + i)^5

now when i do this i always end up getting

-(32squareroot(3))/2 + i32/2

the book seems to get teh same answer except WITHOUT the 2 in the denominator and I have tried this problem over and over always getting the same thing I have a test on this tomarrow and I really would appreciate some explanation of how to get rid of that two

## Answers and Replies

Integral
Staff Emeritus
Science Advisor
Gold Member
Why not show us how you got the 2?

Please show us your work.

k sry

Alright I started by noticing the radius was already in front in this problem (the two out in front) so I moved to the next step putting

2^5 then getting cos squareroot(3)/2 + isin 1/2 which is the same as
cis (Pi/6)
(cis(pi/6) is the same as cos(pi/6) + sin (pi/6) for those of you that don't know)
then i got 2^5 to be 32 so now i had

32(cis(pi/6)) so I multpilied the 5 by pi/6 in coordination with the theorem
getting 32(cis(5pi/6)) then I got cosine of 5pi/6 to be - squareroot(3)/2 and the sin of 5pi/6 to be 1/2 and then finally multiplied these answers by 32 ultimately getting

-32squareroot(3)/2 +i32sin

you factored out a 2 from the cis, in order to get the right values for the sin and cos. But you didn't add it to the powers of two on the left, so it is really 2^6 instead of 2^5. Then when you divide by 2 again you get the 32.

HallsofIvy
Science Advisor
Homework Helper
physstudent1 said:
Alright I started by noticing the radius was already in front in this problem (the two out in front)
?? The number is $2(\sqrt{3}+ i)$ and 2 is not the "radius". $|\sqrt{3}+ i|= 2$ so $|2(\sqrt{3}+ i)|= 4$.

thank you all

?? The number is $2(\sqrt{3}+ i)$ and 2 is not the "radius". $|\sqrt{3}+ i|= 2$ so $|2(\sqrt{3}+ i)|= 4$.

what would the answer in standard from be if the problem was 2(sqrt3 + i)^7

Mentallic
Homework Helper
what would the answer in standard from be if the problem was 2(sqrt3 + i)^7

You should be creating a new thread.

Convert $\sqrt{3}+i$ into modulus-argument form first.