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I am having trouble identify the smallest positive integer $n$ such that $(\frac{1+i}{1-i})^n = 1$

Can someone please throw on approach?

(Also, please correct the equation in the form of Tex/Latex format).

Thanks.

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LaTeX is not introduced, did I interpreted your question appropriately? –  Quixotic Dec 13 '11 at 13:27
    
@MaX: $\left( \frac{1+n}{1-n} \right)^n = 1 $ has no positive integer solutions –  Daniel Freedman Dec 13 '11 at 13:28
    
I would also like to know that many users find questions posted in the imperative ("Show that", "Prove", "Do", "Find") unpleasant and somewhat rude. –  Quixotic Dec 13 '11 at 13:29
    
I am sorry Max, I would just update the post. –  Nikhil Mulley Dec 13 '11 at 13:30
    
This is real-analysis?! –  Quixotic Dec 13 '11 at 13:34
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1 Answer

EDIT: Scrap what's below the line, the original question has been changed several times.

If your questions means "Find the smallest positive integer $n$ such that $\left(\frac{1+i}{1-i}\right)^n = 1$, where $i$ is the imaginary unit", then I would first make the denominator of your fraction real (by multiplying top and bottom by $1+i$). After simplification, we find that we are looking for the smallest $n$ such that $ i^n = 1 $.

Alternatively, as GEdgar suggests above, we could convert to polar coordinates. What moduli do $1+i$ and $1-i$ have? What are their arguments? So what is the modulus and argument of $\frac{1+i}{1-i}$?


If your question meant "Find the smallest positive integer $n$ such that $\left(\frac{1+n}{1-n}\right)^n = 1$", then the answer is "there are no solutions". Clearly we'd need $\frac{1+n}{1-n} = 1$ if this were to work, but then we're forced to have $n = 0$.

If your question meant "Find the smallest positive integer $n$ such that $\left(1+ \frac{n}{1-n}\right)^n = 1$", you can use the same logic as above to arrive at the answer.

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