Key Concepts and Formulas

• Powers of the imaginary unit: $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$. In general, $i^{4k} = 1, i^{4k+1} = i, i^{4k+2} = -1, i^{4k+3} = -i$ where $k$ is an integer.
• Complex number arithmetic: $(a+bi) + (c+di) = (a+c) + (b+d)i$, $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$.
• Condition for a positive integer: A complex number $z = a + bi$ is a positive integer if and only if $b=0$ and $a$ is a positive integer.

Step-by-Step Solution

1. Simplify the expression

We are given the expression $\frac{(2i)^n}{(1-i)^{n-2}}$. Our goal is to simplify this expression and find the smallest positive integer $n$ such that the result is a positive integer.

2. Rewrite the denominator using $(1-i)^2 = -2i$

We know that $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$. This is a crucial identity for simplifying expressions involving $(1-i)$. We rewrite the denominator as follows: $(1-i)^{n-2} = \frac{(1-i)^n}{(1-i)^2} = \frac{(1-i)^n}{-2i}$ Why this step? We can rewrite the denominator in terms of $(1-i)^2$ which simplifies to $-2i$, and also isolate the term $(1-i)^n$.

3. Substitute the rewritten denominator into the original expression

Substituting this back into the original expression, we get: $\frac{(2i)^n}{(1-i)^{n-2}} = \frac{(2i)^n}{\frac{(1-i)^n}{-2i}} = \frac{(2i)^n (-2i)}{(1-i)^n} = (-2i) \frac{(2i)^n}{(1-i)^n} = (-2i) \left(\frac{2i}{1-i}\right)^n$ Why this step? This substitution allows us to group the terms raised to the power of $n$ together.

4. Simplify the fraction $\frac{2i}{1-i}$

To simplify the fraction, we multiply the numerator and denominator by the conjugate of the denominator: $\frac{2i}{1-i} = \frac{2i(1+i)}{(1-i)(1+i)} = \frac{2i + 2i^2}{1 - i^2} = \frac{2i - 2}{1 - (-1)} = \frac{-2 + 2i}{2} = -1 + i$ Why this step? Rationalizing the denominator makes the fraction easier to work with.

5. Rewrite $-1+i$ in terms of $\sqrt{2}e^{i\frac{3\pi}{4}}$

Notice that $(-1+i) = \sqrt{(-1)^2 + 1^2} e^{i \theta} = \sqrt{2} e^{i \frac{3\pi}{4}}$.

6. Rewrite $\frac{(2i)^n}{(1-i)^{n-2}}$

Substituting this result back into the expression, we have: $(-2i)\left(\frac{2i}{1-i}\right)^n = (-2i)(-1+i)^n$ Now, we want to simplify $-2i(-1+i)^n$.

7. Rewrite $-2i$ as $(1-i)^2$

Since $(1-i)^2 = -2i$, we can substitute to get: $(1-i)^2 (-1+i)^n$ We also know that $-1+i = \sqrt{2} e^{i \frac{3\pi}{4}}$, and $1-i = \sqrt{2}e^{-i \frac{\pi}{4}}$. So we have: $(\sqrt{2}e^{-i \frac{\pi}{4}})^2 (\sqrt{2}e^{i \frac{3\pi}{4}})^n = (\sqrt{2})^2 (e^{-i \frac{\pi}{4}})^2 (\sqrt{2})^n (e^{i \frac{3\pi}{4}})^n$ $= 2(\sqrt{2})^n e^{-i \frac{\pi}{2}} e^{i \frac{3n\pi}{4}} = 2(\sqrt{2})^n e^{i (\frac{3n\pi}{4} - \frac{\pi}{2})} = 2(\sqrt{2})^n e^{i (\frac{3n\pi - 2\pi}{4})} = 2(\sqrt{2})^n e^{i \pi (\frac{3n - 2}{4})}$

8. Find conditions on n for the expression to be a positive integer

For this expression to be a positive integer, the imaginary part must be 0, which means that $e^{i \pi (\frac{3n - 2}{4})}$ must be a real number. In order for this to be a real number, $\frac{3n-2}{4}$ must be an integer. Let $\frac{3n-2}{4} = k$ for some integer $k$. Then $3n-2 = 4k$, so $3n = 4k + 2$, which means $n = \frac{4k+2}{3}$. Since $n$ must be an integer, $4k+2$ must be divisible by 3. We can test values of $k$ to find the smallest $k$ such that $4k+2$ is divisible by 3. If $k=0$, $4k+2 = 2$, not divisible by 3. If $k=1$, $4k+2 = 6$, divisible by 3. So, $k=1$ is the smallest integer that works. Therefore, $n = \frac{4(1)+2}{3} = \frac{6}{3} = 2$. Now we need to check the value of $2 (\sqrt{2})^n e^{i \pi (\frac{3n - 2}{4})}$ for $n=2$. $2 (\sqrt{2})^2 e^{i \pi (\frac{3(2) - 2}{4})} = 2(2) e^{i \pi (\frac{4}{4})} = 4 e^{i \pi} = 4(-1) = -4$ Since we want a positive integer, we need to continue testing values for $k$. If $k=4$, $4k+2 = 18$, divisible by 3. So, $k=4$ is the next integer that works. Therefore, $n = \frac{4(4)+2}{3} = \frac{18}{3} = 6$. Now we need to check the value of $2 (\sqrt{2})^n e^{i \pi (\frac{3n - 2}{4})}$ for $n=6$. $2 (\sqrt{2})^6 e^{i \pi (\frac{3(6) - 2}{4})} = 2(8) e^{i \pi (\frac{16}{4})} = 16 e^{i 4\pi} = 16(1) = 16$ Since $16$ is a positive integer, $n=6$ is the smallest positive integer.

9. Simplify using $n=2$

If $n=2$, we have $\frac{(2i)^2}{(1-i)^{2-2}} = \frac{(2i)^2}{(1-i)^0} = (2i)^2 = 4i^2 = -4$, which is an integer but not a positive integer.

10. Simplify using $n=6$ If $n=6$, we have $\frac{(2i)^6}{(1-i)^{6-2}} = \frac{(2i)^6}{(1-i)^4} = \frac{2^6 i^6}{((1-i)^2)^2} = \frac{64(-1)}{(-2i)^2} = \frac{-64}{4i^2} = \frac{-64}{-4} = 16$, which is a positive integer.

Common Mistakes & Tips

• Remember to multiply by the conjugate when dividing complex numbers.
• Keep track of the powers of $i$ and remember their cyclic nature.
• Be careful with signs, especially when dealing with squares and other even powers of negative numbers.

Summary

We simplified the given expression and found that the smallest positive integer $n$ for which the expression is a positive integer is $n=6$.

The final answer is \boxed{6}.