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    1Show that, in general, there are exactly 2 solutions to equation:

    zz*+az*+b*z+c=0

    for a,b,c complex. Alsom z*=complex conjugate of z for clarity.

    2When are there more than 2 solutions/

    How do i do first part..i have a feeling second part should follow on fine
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    (Original post by HeavisideDelts)
    1Show that, in general, there are exactly 2 solutions to equation:

    zz*+az*+b*z+c=0

    for a,b,c complex. Alsom z*=complex conjugate of z for clarity.

    2When are there more than 2 solutions/

    How do i do first part..i have a feeling second part should follow on fine
    



let z = a + bj

\Rightarrow z^* = a - bj

    That is where I would start.
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    (Original post by HeavisideDelts)
    1Show that, in general, there are exactly 2 solutions to equation:

    zz*+az*+b*z+c=0

    for a,b,c complex. Alsom z*=complex conjugate of z for clarity.

    2When are there more than 2 solutions/

    How do i do first part..i have a feeling second part should follow on fine
    Note that the LHS is (z+a)(z^*+b^*)-a b^*+c, so you're trying to solve (z+a)(z^*+b^*) = r for some r \in \mathbb{C}. Does that help?
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    (Original post by lizard54142)
    



let z = a + bj

\Rightarrow z^* = a - bj

    That is where I would start.
    (Original post by Smaug123)
    Note that the LHS is (z+a)(z^*+b^*)-a b^*+c, so you're trying to solve (z+a)(z^*+b^*) = r for some r \in \mathbb{C}. Does that help?
    Sorry i made a mistake in typing up the equation. It was supposed to be:

    zz*+az+b*z+c=0
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    (Original post by lizard54142)
    



let z = a + bj

\Rightarrow z^* = a - bj

    That is where I would start.
    But a,b,c are complex so if i wanted to say set Re=0 and Im=0 then id need to write a,b,c in x+iy form which seems quite cumbersome?
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    (Original post by Smaug123)
    Note that the LHS is (z+a)(z^*+b^*)-a b^*+c, so you're trying to solve (z+a)(z^*+b^*) = r for some r \in \mathbb{C}. Does that help?
    If the question was what i originally put and then id use what you say. How would i go about continuing from there?
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    (Original post by HeavisideDelts)
    But a,b,c are complex so if i wanted to say set Re=0 and Im=0 then id need to write a,b,c in x+iy form which seems quite cumbersome?
    Is this for all a, b, c, z \in \mathbb{C} ?
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    (Original post by lizard54142)
    Is this for all a, b, c, z \in \mathbb{C} ?
    Yep
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    (Original post by HeavisideDelts)
    Yep
    My apologies, I misread the question. That makes things more difficult, I'll let Smaug help you as he is much more qualified than me.
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    (Original post by Smaug123)
    Note that the LHS is (z+a)(z^*+b^*)-a b^*+c, so you're trying to solve (z+a)(z^*+b^*) = r for some r \in \mathbb{C}. Does that help?
    i went through by writing each complex number in x+iy form . Set Real and imaginary parts equal to 0 . This gave me two simultaneous eqns of form:

    x^2+y^2+Ax+By+C=0

    and

    Ay-Bx+D=0

    Substituted for y in terms of x then you will get a quadratic (in x). So say Kx^2+Lx+R=0. Then in general theres two solutions for z=x+iy, (as y is uniquely deteremined from x) and the case of more than two would be where K=L=0?

    Is that correct
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    (Original post by HeavisideDelts)
    If the question was what i originally put and then id use what you say. How would i go about continuing from there?
    We can scale further to get z (z+k)^* = f for some k, by substituting u = z+a.

    That happens to be in the same form as your corrected version, so in fact the two problems are basically equivalent.

    That's scaled out nearly everything there is to scale. We can also scale f down to have modulus 1, by dividing through by |f| (after first doing the case that f=0). Strictly, that's substituting u = \frac{z}{\sqrt{|f|}}, and we again alter k to match.

    That is, we're down to solving z(z+k)^* = e^{i \theta}, where \theta \in [0, 2 \pi ) is fixed and k is fixed but arbitrary.

    I've not got any further than this, I'm afraid, but it seems much more amenable to solution than the original.
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    (Original post by HeavisideDelts)
    i went through by writing each complex number in x+iy form . Set Real and imaginary parts equal to 0 . This gave me two simultaneous eqns of form:

    x^2+y^2+Ax+By+C=0

    and

    Ay-Bx+D=0

    Substituted for y in terms of x then you will get a quadratic (in x). So say Kx^2+Lx+R=0. Then in general theres two solutions for z=x+iy, (as y is uniquely deteremined from x) and the case of more than two would be where K=L=0?

    Is that correct
    That looks reasonable, at least. I'm quite tired today :P
 
 
 
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