Divide Polar Form: Solve 5∠2.214/√5∠-1.107

AI Thread Summary
To divide two complex numbers in polar form, divide their magnitudes and subtract their angles. For the expression (5∠2.214)/(√5∠-1.107), the magnitude simplifies to √5, and the angle is calculated as 2.214 - (-1.107), resulting in an angle of approximately 3.321. There is a concern regarding the angle's quadrant, as the calculator provides an angle of 0.1799, which may indicate a co-terminal angle or a quadrant issue. The discussion highlights the importance of correctly determining the angle's quadrant when using arctan. Overall, the solution is confirmed to be in polar form as required.
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Homework Statement


(5∠2.214)/(√5∠-1.107)
ive gotten this far in a problem(thats the answer but i need to simplify).. all i need to know is how to divide the angles?


Homework Equations





The Attempt at a Solution

 
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pat666 said:

Homework Statement


(5∠2.214)/(√5∠-1.107)
ive gotten this far in a problem(thats the answer but i need to simplify).. all i need to know is how to divide the angles?


Homework Equations





The Attempt at a Solution

Divide the magnitudes and subtract the angles.
 
so that would be sqrt(5) L3.32145... could you take a look at the original problem and tell me if I am right...(1+2j)^2/(1-2j) they wanted the solution in polar.
 
thanks i just wasnt sure about the angle, my calculator says it should be 0.1799, probably co-terminal or in a different quadrant or something?
 
Different quadrant - third quadrant. If you used arctan, you'll get an angle in the first or fourth quadrant.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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