SOLUTION: Find the polar representation of Z= Cosa+Sina+i(Sina-Cosa) where a<[0, 2&#960;)

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Question 773999: Find the polar representation of Z= Cosa+Sina+i(Sina-Cosa) where a<[0, 2π)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
To find the argument theta,

IMAGINARY_PART%2FREAL_PART%22%22=%22%22tan%28theta%29%22%22=%22%22%28sin%28A%29-cos%28A%29%29%2F%28cos%28A%29%2Bsin%28A%29%29

Write tan%28theta%29 as sin%28theta%29%2Fcos%28theta%29

sin%28theta%29%2Fcos%28theta%29%22%22=%22%22%28sin%28A%29-cos%28A%29%29%2F%28cos%28A%29%2Bsin%28A%29%29

Cross-multiply

sin%28theta%29%28cos%28A%29%2Bsin%28A%29%29%22%22=%22%22cos%28theta%29%28sin%28A%29-cos%28A%29%29 

sin%28theta%29cos%28A%29%2Bsin%28theta%29sin%28A%29%22%22=%22%22cos%28theta%29sin%28A%29-cos%28theta%29cos%28A%29 

Rearrange the equation:

sin%28theta%29cos%28A%29-cos%28theta%29sin%28A%29%22%22=%22%22-sin%28theta%29sin%28A%29-cos%28theta%29cos%28A%29 

sin%28theta%29cos%28A%29-cos%28theta%29sin%28A%29%22%22=%22%22-%28sin%28theta%29sin%28A%29%2Bcos%28theta%29cos%28A%29%29 

sin%28theta%29cos%28A%29-cos%28theta%29sin%28A%29%22%22=%22%22-%28cos%28theta%29cos%28A%29%2Bsin%28theta%29sin%28A%29%29  

sin%28theta-A%29%22%22=%22%22-cos%28theta-A%29

Divide both sides by cos%28theta-A%29

sin%28theta-A%29%2Fcos%28theta-A%29%22%22=%22%22-1

tan%28theta-A%29%22%22=%22%22-1

theta-A%22%22=%22%223pi%2F4, 7pi%2F4

theta%22%22=%22%223pi%2F4%2BA, 7pi%2F4%2BA

To find the modulus or absolute value r:

r%5E2%22%22=%22%22%28REAL_PART%29%5E2%2B%28IMAGINARY_PART%29%5E2

r%5E2%22%22=%22%22%28cos%28A%29%2Bsin%28A%29%29%5E2%22%22%2B%22%22%28sin%28A%29-cos%28A%29%29%5E2

r%5E2%22%22=%22%22

Using the identity cosineČ + sineČ = 1, the right side is just 2

rČ = 2

r = √2 

Polar forms:

√2[cos(3pi%2F4%2BA) + i·sin(3pi%2F4%2BA)]

√2[cos(7pi%2F4%2BA) + i·sin(7pi%2F4%2BA)]

Edwin