\left(\frac{1}{\sqrt{2}\:}-\frac{i}{\sqrt{2}}\right)^{70}

To evaluate the expression 

\[
\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)^{70}
\]

we can use the polar (or exponential) form of the complex number.

1. **Rewrite the complex number in polar form**:
   
   The expression \(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\) can be rewritten as:
   
   \[
   z = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}
   \]

   This complex number has a modulus:
   
   \[
   |z| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1
   \]

   Since the modulus is \(1\), the magnitude of this complex number does not change with powers.

2. **Find the argument of \(z\)**:

   The argument (or angle \(\theta\)) of \(z\) is calculated by:
   
   \[
   \theta = \arctan\left(\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right) = \arctan(-1) = -\frac{\pi}{4}
   \]

   Therefore, we can rewrite \(z\) in polar form as:
   
   \[
   z = e^{-i\frac{\pi}{4}}
   \]

3. **Raise \(z\) to the 70th power**:

   Now, we calculate \(z^{70}\) as:
   
   \[
   z^{70} = \left(e^{-i\frac{\pi}{4}}\right)^{70} = e^{-i\frac{70\pi}{4}} = e^{-i\frac{35\pi}{2}}
   \]

4. **Simplify the exponent**:

   To simplify \(e^{-i\frac{35\pi}{2}}\), we can reduce the angle by recognizing that any integer multiple of \(2\pi\) does not affect the result due to periodicity. So:
   
   \[
   -\frac{35\pi}{2} = -18\pi - \frac{\pi}{2} = -9(2\pi) - \frac{\pi}{2} \equiv -\frac{\pi}{2} \pmod{2\pi}
   \]

   Thus:
   
   \[
   e^{-i\frac{35\pi}{2}} = e^{-i\frac{\pi}{2}}
   \]

5. **Express the result in rectangular form**:

   Now we evaluate \(e^{-i\frac{\pi}{2}}\), which corresponds to \(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\):
   
   \[
   e^{-i\frac{\pi}{2}} = 0 - i = -i
   \]

Therefore, the final result is:

\[
\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)^{70} = -i
\]

---

Would you like further details, or do you have any questions?

Here are some related questions to deepen understanding:

1. How do you convert a complex number to polar form?
2. Why does raising a complex number to a power involve multiplying the argument by the power?
3. What is the significance of the modulus of a complex number being \(1\) in exponentiation?
4. How can the periodicity of \(2\pi\) in the argument affect complex powers?
5. How does Euler's formula help in simplifying powers of complex numbers?

**Tip**: Always check the modulus when dealing with complex powers, as a modulus of \(1\) simplifies calculations significantly!

\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)^{70}

Explore the step-by-step solution to evaluate the complex number (1/√2 - i/√2)^70 using polar form and Euler's formula. Learn about modulus, argument, and the impact of periodicity in complex powers. Suitable for advanced high school students (Grades 11-12).

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