$\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)^9$

To solve $\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)^9$, we can use polar form and properties of complex numbers. Here’s a step-by-step process:

1. **Simplify the fraction:**
   \[
   \frac{1-\mathrm{i}}{1+\mathrm{i}}
   \]
   To simplify, multiply the numerator and denominator by the conjugate of the denominator:
   \[
   \frac{(1-\mathrm{i})(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})}
   \]
   The denominator becomes:
   \[
   (1+\mathrm{i})(1-\mathrm{i}) = 1 - \mathrm{i}^2 = 1 - (-1) = 2
   \]
   The numerator becomes:
   \[
   (1-\mathrm{i})^2 = 1 - 2\mathrm{i} + \mathrm{i}^2 = 1 - 2\mathrm{i} - 1 = -2\mathrm{i}
   \]
   Thus, we have:
   \[
   \frac{-2\mathrm{i}}{2} = -\mathrm{i}
   \]

2. **Raise to the 9th power:**
   Now we need to find $(-i)^9$.

   $-\mathrm{i}$ in polar form is $\mathrm{e}^{-\mathrm{i}\pi/2}$ because $-\mathrm{i}$ lies on the negative imaginary axis with an argument of $-\pi/2$.

   So:
   \[
   (-\mathrm{i})^9 = \left(\mathrm{e}^{-\mathrm{i}\pi/2}\right)^9 = \mathrm{e}^{-\mathrm{i}9\pi/2}
   \]

3. **Simplify the exponent:**
   The argument $-9\pi/2$ can be simplified modulo $2\pi$:
   \[
   -9\pi/2 = -4\pi - \pi/2 = -2(2\pi) - \pi/2
   \]
   Since $\mathrm{e}^{-2\pi\mathrm{i}k} = 1$ for any integer $k$, we can ignore the multiple of $2\pi$:
   \[
   \mathrm{e}^{-\mathrm{i}9\pi/2} = \mathrm{e}^{-\mathrm{i}\pi/2}
   \]
   \[
   \mathrm{e}^{-\mathrm{i}\pi/2} = -\mathrm{i}
   \]

Therefore:
\[
\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)^9 = -\mathrm{i}
\]

Would you like more details or have any questions about this solution? Here are five more questions you might be interested in:

1. How do you convert complex numbers to polar form?
2. What are the properties of raising complex numbers to a power?
3. How do you multiply and divide complex numbers?
4. What is Euler's formula and how is it used in complex numbers?
5. How do you simplify complex fractions?

**Tip:** When dealing with complex numbers, converting them to polar form can simplify many operations like multiplication, division, and exponentiation.

Learn how to solve the problem $\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)^9$ using properties of complex numbers and exponentiation. This problem involves converting complex numbers to polar form, applying Euler's formula, and simplifying exponents. Suitable for advanced high school students.

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