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

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Homework Help Overview

The discussion revolves around determining the values of \( m \) for which the complex number \( (\sqrt{3}+i)^m \) is positive and real. The problem involves complex numbers and their properties, particularly in exponential form.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the conversion of complex numbers into exponential form and discuss conditions under which two complex expressions are equal. There is an attempt to derive values for \( n \) based on angle equivalences and periodicity in complex exponentials.

Discussion Status

Some participants have provided hints and guidance on simplifying the problem and questioning the assumptions made regarding angles. There is an ongoing exploration of the relationship between the expressions and the conditions for positivity and reality of the complex number.

Contextual Notes

There is mention of potential discrepancies between calculated values and those provided in a reference, indicating a need for careful consideration of the angles involved. Participants are also discussing the implications of rewriting the expressions in different forms.

Santilopez10
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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!
 
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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?
 

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