For what m is the complex number $(\sqrt 3+i)^m$ positive and real?

AI Thread Summary
The discussion focuses on finding integer values of n for which the equation (\sqrt{3}+i)^n = 2^{n-1}(-1+\sqrt{3}i) holds true. The conversion to complex exponential form reveals that the angles must match, leading to the equation n(\pi/3) = 2\pi/3, which suggests potential solutions like n = 2, 8, 14, etc. However, a miscalculation regarding the angle for \sqrt{3}+i is identified, correcting it to 2e^{(\pi/3)i}. The final goal is to determine for which m the expression (\sqrt{3}+i)^m is positive and real, with the correct interpretation of the equation leading to insights about the periodic nature of complex exponentials.
Santilopez10
Messages
81
Reaction score
8

Homework Statement


Find all $$n \in Z$$, for which $$ (\sqrt 3+i)^n = 2^{n-1} (-1+\sqrt 3 i)$$

Homework Equations


$$ (a+b i)^n = |a+b i|^n e^{i n (\theta + 2 \pi k)} $$

The Attempt at a Solution


First I convert everything to it`s complex exponential form: $$ 2^n e^{i n (\frac {\pi}{3}+ 2\pi k)} = 2^{n-1} 2 e^{i (\frac{2 \pi}{3} +2 \pi k)} $$
this simplifies to $$ e^{i n (\frac {\pi}{3}+ 2\pi k)} = e^{i (\frac{2 \pi}{3} +2 \pi k)} $$
I know how to find an expression for n, but not that it`s only in the field of ## Z ##, any help would be appreciated, thanks!
 
Physics news on Phys.org
Hint: There is no point to writing out the ks. Instead, try to figure out when the numbers are equal without using them. Also, if you do use them, you cannot assume that they are the same on both sides.
 
Well I would get $$ n \frac {\pi}{3} = \frac {2 \pi}{3} $$
which is true for n = 2, n =8, n= 14, n= 20 ... 2+6t if I am not wrong, but this still does not satisfy the answer provided by the book which is n= 4(1+3t) I am missing a 2!
 
Santilopez10 said:
Well I would get $$ n \frac {\pi}{3} = \frac {2 \pi}{3} $$
which is true for n = 2, n =8, n= 14, n= 20 ... 2+6t if I am not wrong, but this still does not satisfy the answer provided by the book which is n= 4(1+3t) I am missing a 2!
Actually that equation is only true for n = 2, but that's beside the point.Your primary difficulty lies in the fact that the ## \ \dfrac \pi 3 \ ## is incorrect.

## \sqrt{3} + i \ne 2 e^{(\pi/3)i} ##

Rather: ##\ 2 e^{(\pi/3)i} = 1 +i\sqrt{3} \,.##
 
SammyS said:
Actually that equation is only true for n = 2, but that's beside the point.Your primary difficulty lies in the fact that the ## \ \dfrac \pi 3 \ ## is incorrect.

## \sqrt{3} + i \ne 2 e^{(\pi/3)i} ##

Rather: ##\ 2 e^{(\pi/3)i} = 1 +i\sqrt{3} \,.##
Thanks, now I got the correct answer. By the way, when I mentioned the various answers for n it was in the context of angles, not numbers a s a whole.
 
I would rewrite the right hand side
$$(\sqrt 3+i)^n = 2^{n-1} (-1+\sqrt 3 i)$$
$$(\sqrt 3+i)^n = 2^{n-4}(\sqrt 3+i)^4$$
for what m is
$$(\sqrt 3+i)^m$$
positive and real?
 
Back
Top