Change the following complex numbers into polar form
`(1+ 3i)/(1-2i)`

Answer

To convert the complex number \((1 + 3i)/(1 - 2i)\) into polar form, we will follow these steps:

### Step 1: Simplify the Complex Number
Let \( z = \frac{1 + 3i}{1 - 2i} \). To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator, which is \( 1 + 2i \).

\[ z = \frac{(1 + 3i)(1 + 2i)}{(1 - 2i)(1 + 2i)} \]

### Step 2: Expand the Numerator
Now, we will expand the numerator:

\[ (1 + 3i)(1 + 2i) = 1 \cdot 1 + 1 \cdot 2i + 3i \cdot 1 + 3i \cdot 2i \]
\[ = 1 + 2i + 3i + 6i^2 \]
Since \( i^2 = -1 \), we have:
\[ = 1 + 5i - 6 = -5 + 5i \]

### Step 3: Expand the Denominator
Now, we will expand the denominator:

\[ (1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5 \]

### Step 4: Combine the Results
Now we can combine the results from the numerator and denominator:

\[ z = \frac{-5 + 5i}{5} = -1 + i \]

### Step 5: Find the Modulus
Next, we need to find the modulus of \( z \):

\[ |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 6: Find the Argument
Now we will find the argument \( \theta \) of \( z \). The argument is given by:

\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{1}{-1}\right) \]

This gives us \( \theta = \tan^{-1}(-1) \), which corresponds to \( \frac{3\pi}{4} \) since \( z \) is in the second quadrant (where \( x < 0 \) and \( y > 0 \)).

### Step 7: Write in Polar Form
Finally, we can write the polar form of \( z \):

\[ z = |z| \left( \cos \theta + i \sin \theta \right) \]
\[ = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) \]

### Final Answer
Thus, the polar form of the complex number \((1 + 3i)/(1 - 2i)\) is:

\[ z = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) \]

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