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Math Help - Vectors + COmplex numbers

  1. #1
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    Vectors + COmplex numbers

    Hi can you help me with:

    Suppose z = cis A where A is acute. Find the argument of z + 1:

    I understand how to draw the vectors and all on the Argand diagram, so the question with the answers and all are in the first image.


    However,t o find the argument of z+1, I don't understand, my understanding is that the argument of z+1 is the red angle, as shown in the next picture. Why is it wrong? For the answer is that the argument of z+1 is pi/2 since it is a rhombus, but I don't see how, not from my understanding of what a z+1 argument is!

    Please help, thank you so much!
    Thank you!
    Attached Thumbnails Attached Thumbnails Vectors + COmplex numbers-zix.png   Vectors + COmplex numbers-zic-2.png  
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  2. #2
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    Re: Vectors + COmplex numbers

    For the answer is that the argument of z+1 is pi/2
    Was this a typo or is this what you understood. The argument is theta/2 not pi/2
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  3. #3
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    Re: Vectors + COmplex numbers

    assuming z = cis(θ) is in the first quadrant, we have arg(z) = arctan(sin(θ)/cos(θ)) = arctan(tan(θ)) = θ.

    now arg(z+1) = arctan(sin(θ)/(1+cos(θ))

    let's look closer at what we're taking the arctangent OF:

    sin(θ)/(1+cos(θ)) = sin(θ/2 + θ/2)/[(1 + cos(θ/2 + θ/2)] = (2sin(θ/2)cos(θ/2))/[1 + cos2(θ/2) - sin2(θ/2)]

    = (2sin(θ/2)cos(θ/2))/(2cos2(θ/2)) = sin(θ/2)/cos(θ/2) = tan(θ/2).

    hence arg(z+1) = arctan(sin(θ)/(1+cos(θ)) = arctan(tan(θ/2)) = θ/2.

    the magnitude is also easy to find:

    |z+1| = √[(1+cos(θ))2 + sin2(θ)] = √[1 + 2cos(θ) + cos2(θ) + sin2(θ)]

    = √(2 + 2cos(θ)) = √(2 + 2cos(θ/2 + θ/2)) = √(2 + 2cos2(θ/2) - 2sin2(θ/2)) = √(4cos2(θ/2))

    = 2cos(θ/2) (because we are in the first quadrant).

    the "geometric" argument is a bit more elegant, and visual, to boot, taking advantage of the symmetry of z and 1 with respect to z+1. this only works because z is the same magnitude as 1.
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  4. #4
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    Re: Vectors + COmplex numbers

    Hi thank you so much!

    How did you know that arg(z+1) = arctan(sin(θ)/(1+cos(θ)), why is the one added to the cos and not the sin etc?

    Thank you!
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    Re: Vectors + COmplex numbers

    the "c" in "cis" stands for cosine, which is the real part of the number cos(θ) + i sin(θ). the number 1 is real, so we add it to the real part.
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  6. #6
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    Re: Vectors + COmplex numbers

    Quote Originally Posted by Tutu View Post
    [COLOR=#1a1a18]Suppose z = cis A where A is acute. Find the argument of z + 1

    There is a very easy and intuitive geometric proof for this.

    The figure OCBA is a rhombus, because z is on the unit circle, |z|=1. As vectors z+1 is a diagonal of that rhombus which bisects the angle that is \text{arg}(z).
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  7. #7
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    Re: Vectors + COmplex numbers

    Quote Originally Posted by Plato View Post
    There is a very easy and intuitive geometric proof for this.

    The figure OCBA is a rhombus, because z is on the unit circle, |z|=1. As vectors z+1 is a diagonal of that rhombus which bisects the angle that is \text{arg}(z).
    yes! such a proof is depicted in the attachment of the original post (the blue diagram).
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