To convert the complex number [tex]\( z = \frac{3\sqrt{3}}{2} - \frac{3}{2}i \)[/tex] to its polar form, we need to find its magnitude and argument.
### Step 1: Calculate the Magnitude
The magnitude [tex]\( |z| \)[/tex] of a complex number [tex]\( z = a + bi \)[/tex] is given by the formula:
[tex]\[ |z| = \sqrt{a^2 + b^2} \][/tex]
Here,
[tex]\[ a = \frac{3\sqrt{3}}{2} \quad \text{(real part)} \][/tex]
[tex]\[ b = -\frac{3}{2} \quad \text{(imaginary part)} \][/tex]
So,
[tex]\[ |z| = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + \left(-\frac{3}{2}\right)^2} \][/tex]
Calculating each term,
[tex]\[ \left(\frac{3\sqrt{3}}{2}\right)^2 = \frac{(3\sqrt{3})^2}{4} = \frac{27}{4} \][/tex]
[tex]\[ \left(-\frac{3}{2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \][/tex]
Adding these,
[tex]\[ |z| = \sqrt{\frac{27}{4} + \frac{9}{4}} = \sqrt{\frac{36}{4}} = \sqrt{9} = 3 \][/tex]
Thus, the magnitude is:
[tex]\[ |z| = 3 \][/tex]
### Step 2: Calculate the Argument
The argument [tex]\( \theta \)[/tex] of a complex number [tex]\( z = a + bi \)[/tex] is given by:
[tex]\[ \theta = \operatorname{atan2}(b, a) \][/tex]
Using [tex]\( a = \frac{3\sqrt{3}}{2} \)[/tex] and [tex]\( b = -\frac{3}{2} \)[/tex],
[tex]\[ \theta = \operatorname{atan2}\left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right) \][/tex]
This yields,
[tex]\[ \theta = -0.5235987755982988 \, \text{(radians)} \][/tex]
### Step 3: Express in Polar Form
A complex number in polar form is expressed as:
[tex]\[ z = r (\cos \theta + i \sin \theta) \][/tex]
Substituting the values we found,
[tex]\[ r = 3 \][/tex]
[tex]\[ \theta = -0.5235987755982988 \][/tex]
So the polar form of [tex]\( z \)[/tex] is:
[tex]\[ z = 3 \left( \cos(-0.5235987755982988) + i \sin(-0.5235987755982988) \right) \][/tex]
Or alternatively, recognizing the angle in radians:
[tex]\[ z = 3 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) \][/tex]
Thus, the complex number [tex]\( z = \frac{3\sqrt{3}}{2} - \frac{3}{2}i \)[/tex] in polar form is:
[tex]\[ z = 3 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) \][/tex]
