To convert a complex number from rectangular form [tex]\( z = x + yi \)[/tex] to polar form [tex]\( r \left( \cos(\theta) + i \sin(\theta) \right) \)[/tex], we need to find the modulus [tex]\( r \)[/tex] and the argument [tex]\( \theta \)[/tex].
Given the complex number [tex]\( z = \frac{5 \sqrt{3}}{4} - \frac{5}{4} i \)[/tex]:
1. Find the Real and Imaginary Parts:
- Real part: [tex]\( x = \frac{5 \sqrt{3}}{4} \)[/tex]
- Imaginary part: [tex]\( y = -\frac{5}{4} \)[/tex]
2. Calculate the Modulus [tex]\( r \)[/tex]:
The modulus [tex]\( r \)[/tex] is given by:
[tex]\[
r = \sqrt{x^2 + y^2}
\][/tex]
Substituting [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
r = \sqrt{\left( \frac{5 \sqrt{3}}{4} \right)^2 + \left( -\frac{5}{4} \right)^2} = \sqrt{\frac{75}{16} + \frac{25}{16}} = \sqrt{\frac{100}{16}} = \sqrt{6.25} = 2.5
\][/tex]
3. Calculate the Argument [tex]\( \theta \)[/tex]:
The argument [tex]\( \theta \)[/tex] is found using the arctangent function:
[tex]\[
\theta = \tan^{-1} \left( \frac{y}{x} \right)
\][/tex]
Substituting [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
\theta = \tan^{-1} \left( \frac{-\frac{5}{4}}{\frac{5 \sqrt{3}}{4}} \right) = \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) = -\frac{\pi}{6}
\][/tex]
Since the argument is negative, we adjust it to the correct quadrant for a complete understanding. This is done by adding [tex]\( 2\pi \)[/tex]:
[tex]\[
\theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}
\][/tex]
4. Expressing the Complex Number in Polar Form:
Combine the modulus [tex]\( r \)[/tex] with the argument [tex]\( \theta \)[/tex]:
[tex]\[
z = r (\cos(\theta) + i \sin(\theta))
\][/tex]
Substituting [tex]\( r = 2.5 \)[/tex] and [tex]\( \theta = \frac{11\pi}{6} \)[/tex]:
[tex]\[
z = 2.5 \left( \cos \left( \frac{11\pi}{6} \right) + i \sin \left( \frac{11\pi}{6} \right) \right)
\][/tex]
Considering the given multiple-choice options:
[tex]\[ \boxed{z = \frac{5}{2}\left(\cos \left(\frac{11 \pi}{6}\right)+i \sin \left(\frac{11 \pi}{6}\right)\right)} \][/tex]
This choice correctly represents the polar form of the given complex number.
