The polar form of the complex number `(i^(25))^(3)` is

Options

A. `cos.(pi)/(3)-isin.(pi)/(3)`
B. `cos(-(pi)/(2)+isin(-(pi)/(2))`
C. `cos.(pi)/(6)-isin.(pi)/(6)`
D. `cos.(pi)/(6)+isin.(pi)/(6)`

Answer

To find the polar form of the complex number \((i^{25})^{3}\), we will follow these steps:

### Step 1: Simplify the expression using exponent rules
We start with the expression:
\[ (i^{25})^{3} \]
Using the property of exponents \((a^m)^n = a^{m \cdot n}\), we can rewrite this as:
\[ i^{25 \cdot 3} = i^{75} \]

**Hint:** Remember that when you raise a power to another power, you multiply the exponents.

### Step 2: Break down the exponent
Next, we can express \(i^{75}\) in a more manageable form. We can break \(75\) into \(72 + 3\):
\[ i^{75} = i^{72 + 3} = i^{72} \cdot i^{3} \]

**Hint:** When dealing with powers of \(i\), breaking the exponent into smaller parts can help simplify calculations.

### Step 3: Use the periodic property of \(i\)
The powers of \(i\) are periodic with a cycle of 4:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

Since \(72\) is a multiple of \(4\), we have:
\[ i^{72} = (i^4)^{18} = 1^{18} = 1 \]
Thus,
\[ i^{75} = 1 \cdot i^{3} = i^{3} \]

**Hint:** Recognizing the periodic nature of \(i\) can simplify your calculations significantly.

### Step 4: Evaluate \(i^{3}\)
From our earlier breakdown, we know:
\[ i^{3} = -i \]

**Hint:** Always refer back to the basic powers of \(i\) to find values quickly.

### Step 5: Convert to polar form
The complex number \(-i\) can be expressed in polar form. Recall that the polar form of a complex number \(z\) is given by:
\[ z = r(\cos \theta + i \sin \theta) \]
where \(r\) is the modulus and \(\theta\) is the argument.

1. **Calculate the modulus \(r\)**:
\[ r = |z| = \sqrt{x^2 + y^2} \]
For \(-i\), we have \(x = 0\) and \(y = -1\):
\[ r = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1 \]

2. **Determine the argument \(\theta\)**:
The argument for \(-i\) is \(-\frac{\pi}{2}\) (since it lies on the negative imaginary axis).

**Hint:** The modulus gives you the distance from the origin, while the argument gives you the angle from the positive real axis.

### Step 6: Write the polar form
Now, substituting \(r\) and \(\theta\) into the polar form:
\[ -i = 1 \left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) \]

Thus, the polar form of the complex number \((i^{25})^{3}\) is:
\[ 1 \left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) \]

**Final Answer:**
\[ 1 \left( \cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right) \right) \]