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

In summary, the equations state that if a and b are real numbers and i is an imaginary number, then their sum, which is also a real number, is equal to the square of i, plus i.
  • #1
Santilopez10
81
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!
 
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  • #2
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.
 
  • #3
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!
 
  • #4
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} \,.##
 
  • #5
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.
 
  • #6
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?
 

1. What is De Moivre's theorem problem?

De Moivre's theorem problem is a mathematical problem that involves finding the nth root of a complex number, where n is a positive integer. It is named after the French mathematician Abraham de Moivre.

2. What is the formula for De Moivre's theorem problem?

The formula for De Moivre's theorem problem is (cosθ + isinθ)^n = cos(nθ) + isin(nθ), where θ is the angle of the complex number and n is the exponent.

3. How is De Moivre's theorem problem used in mathematics?

De Moivre's theorem problem is used to simplify complex number calculations and to find roots of complex numbers. It is also used in trigonometry to find the nth roots of trigonometric functions.

4. What are the applications of De Moivre's theorem problem?

De Moivre's theorem problem has applications in various fields such as engineering, physics, and computer science. It is used to solve problems involving complex numbers, such as in circuit analysis, signal processing, and quantum mechanics.

5. Are there any limitations to De Moivre's theorem problem?

One limitation of De Moivre's theorem problem is that it can only be applied to complex numbers in polar form. It cannot be used for complex numbers in rectangular form. Additionally, the theorem only works for integer exponents and cannot be applied to fractional or negative exponents.

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