What is the polar form of the complex number 3-4i

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SUMMARY

The polar form of the complex number 3-4i is represented as 5(cos(arctan(-4/3)) + i*sin(arctan(-4/3))). However, the quiz incorrectly presented the answer as 5(3/5 - i*4/5), which does not maintain the polar form since it evaluates the trigonometric functions. The correct polar representation retains the angle theta, which is derived from the 3-4-5 right triangle, where cos(theta) = 3/5 and sin(theta) = -4/5. The confusion arises from the interpretation of polar form, which should ideally include the angle rather than its evaluated sine and cosine values.

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  • Knowledge of polar coordinates and trigonometric functions
  • Familiarity with the concept of arctangent in relation to angles
  • Basic geometry involving right triangles, specifically the 3-4-5 triangle
NEXT STEPS
  • Study the derivation of polar form for complex numbers
  • Learn about the differences between polar form and rectangular form
  • Explore the use of the cis function and Euler's formula in complex analysis
  • Investigate common misconceptions in interpreting polar coordinates
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Students studying complex numbers, educators teaching trigonometry and complex analysis, and anyone seeking to clarify the concept of polar form in mathematics.

willibre
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Homework Statement


What is the polar form of the complex number 3-4i?


Homework Equations



z=r*cos(theta)+i*r*sin(theta)

The Attempt at a Solution


5(cos(arctan(-4/3))-i*sin(arctan(-4/3)))

This is what I thought the correct answer would be, but it was a multiple choice quiz and this was not one of the options. The correct answer, according to the quiz, was 5(3/5-i*4/5). I don't see how that is in polar form, since the cos and sin of theta were both evaulated. If that is actually correct, then couldn't you simply make any standard complex number polar form by factoring out r?
 
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For the angle theta in question, the triangle is a 3-4-5 right triangle. cos(theta) = 3/5 and sin(theta) = -4/5.
 
willibre said:

Homework Statement


What is the polar form of the complex number 3-4i?


Homework Equations



z=r*cos(theta)+i*r*sin(theta)

The Attempt at a Solution


5(cos(arctan(-4/3))-i*sin(arctan(-4/3)))

This is what I thought the correct answer would be, but it was a multiple choice quiz and this was not one of the options. The correct answer, according to the quiz, was 5(3/5-i*4/5). I don't see how that is in polar form, since the cos and sin of theta were both evaulated. If that is actually correct, then couldn't you simply make any standard complex number polar form by factoring out r?

Except for the arithmetic mistake already pointed out, I agree with you. Some authors might call the polar form r\, cis(\theta) or re^{i\theta} which would make the argument even stronger that their "best" answer isn't appropriate.
 
I guess my question is, why is it still considered polar form if you carry out the evaluation of cos and sin, thus removing theta? Why wouldn't any polar form then just be r(a/r+ib/r). I guess I don't understand what is technically considered polar form.
 
willibre said:
I guess my question is, why is it still considered polar form if you carry out the evaluation of cos and sin, thus removing theta?

It isn't. The answers to that test question were incorrectly presented.
 

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