Convert the following in polar form,
(i)`-3sqrt2+3sqrt2i`

Answer

To convert the complex number \(-3\sqrt{2} + 3\sqrt{2}i\) into polar form, we will follow these steps:

### Step 1: Identify \(a\) and \(b\)
The complex number can be expressed in the form \(a + bi\), where:
- \(a = -3\sqrt{2}\)
- \(b = 3\sqrt{2}\)

### Step 2: Calculate the modulus \(r\)
The modulus \(r\) of the complex number is given by the formula:
\[ r = \sqrt{a^2 + b^2} \]
Calculating \(a^2\) and \(b^2\):
\[ a^2 = (-3\sqrt{2})^2 = 9 \cdot 2 = 18 \]
\[ b^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18 \]
Now, substituting these values into the modulus formula:
\[ r = \sqrt{18 + 18} = \sqrt{36} = 6 \]

### Step 3: Calculate the argument \(\theta\)
The argument \(\theta\) can be found using the formulas:
\[ \cos \theta = \frac{a}{r} \quad \text{and} \quad \sin \theta = \frac{b}{r} \]
Calculating \(\cos \theta\):
\[ \cos \theta = \frac{-3\sqrt{2}}{6} = -\frac{\sqrt{2}}{2} \]
Calculating \(\sin \theta\):
\[ \sin \theta = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2} \]
Now, we need to find \(\theta\) such that:
- \(\cos \theta = -\frac{\sqrt{2}}{2}\)
- \(\sin \theta = \frac{\sqrt{2}}{2}\)

The angle that satisfies these conditions is:
\[ \theta = \frac{3\pi}{4} \quad \text{(in the second quadrant)} \]

### Step 4: Write in polar form
Now that we have \(r\) and \(\theta\), we can express the complex number in polar form:
\[ z = r(\cos \theta + i \sin \theta) \]
Substituting the values we found:
\[ z = 6\left(\cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4}\right) \]

### Final Answer
Thus, the polar form of the complex number \(-3\sqrt{2} + 3\sqrt{2}i\) is:
\[ z = 6\left(\cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4}\right) \]