Argument value in complex numbers in polar form.

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kayanian
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-1-(under root)3 i
here we find that
r=2(hypotenuse)
a=-1(base)
b=-(under root)3

when i take sin theat= p/h=-(under root)3 / 2
theat from sin is -60

when i take cos theta = b/h =-1 / 2
which gives 120

now one is -60 and other is 120, which is the angel , i have to follow and what do i add it to?
 
Attachments
  • Textbook page titled EXERCISE 1.1 (Page 20) showing a worked solution converting -√3 + i to polar form, with r = 2 and arg z = 5π/6 handwritten-style printed working.
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  • Printed textbook page titled Exercise 1.1 (Page 20) showing a worked solution for converting complex numbers to polar form, with cos θ = -√3/2 and sin θ = 1/2 marked, and a brace noting θ lies in the second quadrant.
    Picture 002.jpg
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  • Handwritten complex number working: z = -√3 + i, r = |z| = 2, cos θ = -√3/2, sin θ = 1/2, θ lies in the second quadrant, Arg z = 5π/6, and z = 2[cos(5π/6 + 2kπ) + i sin(5π/6 + 2kπ)] = 2 cis 5π/6.
    Picture 003.jpg
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Last edited:
Think about where [itex]-1- \sqrt{3}i[/itex] is on the complex plane. It has negative real part and negative imaginary part so it is in the third quadrant. Neither -60 or 120 degrees is in the third quadrant- they are both wrong.

tan(theta)= -sqrt{3}/-1= sqrt(3) gives theta= 60 degrees but since this number is in the third quadrant, theta= 180+ 60= 240 degrees.

cos(240)= -1/2 and sin(210)= -sqrt(3)/2.
 
HallsofIvy said:
Think about where [itex]-1- \sqrt{3}i[/itex] is on the complex plane. It has negative real part and negative imaginary part so it is in the third quadrant. Neither -60 or 120 degrees is in the third quadrant- they are both wrong.

tan(theta)= -sqrt{3}/-1= sqrt(3) gives theta= 60 degrees but since this number is in the third quadrant, theta= 180+ 60= 240 degrees.

cos(240)= -1/2 and sin(210)= -sqrt(3)/2.

So the answer is :

2{cos 120+ i sin 120)
= 2 cis 120
??

but my solution manual gave me this answer

2 Cis 210
which i myself think is wrong

the table of cos and sin positive value is also used if i m right.and when cos and sin are yielding different results, we go to tangent.ok i get it now but i m lil confused because my solution manual gave wrong answer so please confirm.
 
Attachments
  • Table with columns cos θ, sin θ and θ listing sign combinations: +ve/+ve gives α (say); +ve/-ve gives -α (say); -ve/+ve gives (π - α); -ve/-ve gives -(π - α).
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  • Handwritten working for a complex number in polar form: r = |z| = sqrt((-1)^2 + (-sqrt3)^2) = 2, with cos θ = -1/2 and sin θ = -sqrt3/2, leading to θ = -2π/3 in the third quadrant.
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  • Blurry handwritten-style printed working on complex numbers: z = -1 - sqrt(3)i, |z| = sqrt((-1)^2 + (-sqrt(3))^2), arg z found from cos θ = -1/2 and sin θ = -sqrt(3)/2, in the third quadrant, Arg z = -2π/3.
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2cis210 is incorrect. As HallsofIvy mentioned the correct argument is 240.

--Elucidus