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Math Help - How to plot these in complex plane? Please help?

  1. #1
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    How to plot these in complex plane? Please help?

    |z+1| = |z-1|

    Im (z^2) = 2

    Re z > -1

    Re (1/z) < 1

    Please just tell me what the graphs of these functions are and how would I know by looking at a function and telling how the graph is going to be

    like for instance 0 < |z-1| < 1 is an annulus so what about the functions above
    please help
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  2. #2
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    Re: How to plot these in complex plane? Please help?

    Quote Originally Posted by szak1592 View Post
    |z+1| = |z-1|

    The above is the set of all points which are equally distance from 1 to -1.
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  3. #3
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    Re: How to plot these in complex plane? Please help?

    thanks and what about the others
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    Re: How to plot these in complex plane? Please help?

    Szak1952

    you must find the locus that each equation represents .

    I will give you an example : Im(z^2) =2....do you understand what does it mean? it simply says that the imaginary part of an unknown complex number z^2 = (x+yi)^2
    is equal to 2 . You must therefore find this imaginary part and put it equal to 2....lets do it. the z^2 =( x+yi)^2 yields after binomial aepansion that
    z^2 = (x^2-y^2 )+(2xy)i. therefore the Im(Z^2) is 2xy and this must be equal to 2
    therefore we have 2xy=2 or xy=1
    now as you probably understand you must plot the function y =1/x this is of course a hyperbola known as reciprocal function...
    check here Graphs of Reciprocal Functions (with worked solutions & videos)
    to see its plot .

    good luck with the others
    MINOAS
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  5. #5
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    Re: How to plot these in complex plane? Please help?

    If z= x+ iy then z^2= x^2- y^2+ 2xyi. Im(z^2)= 2xy= 2. Graph that.

    If z= x+ iy then Re(z)= x> -1

    If z= x+ iy then \frac{1}{z}= \frac{1}{x+ iy}\frac{x- iy}{x- iy}= \frac{x- iy}{x^2+ y^2} so that Re(1/z)= \frac{x}{x^2+ y^2}< 1

    The best way to graph an inequality is to first graph the associated equality which is the boundary of the graph of the inequality.
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