Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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.
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.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.