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    z^2 + 25 = 0

    not sure about my answer.
    i have formed 2 eq x^2 + y^2 + 25 = 0 and 2xy = 0
    by sub of z = x+yi

    im not sure if i can solve from here since i would need to divide by x and y.

    could i do z = sqrt 25 giving (x+yi) = 5

    y is then = 0 and x = 5?
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    (Original post by YaYaD)
    z^2 + 25 = 0

    not sure about my answer.
    i have formed 2 eq x^2 + y^2 + 25 = 0 and 2xy = 0
    by sub of z = x+yi
    You've lost me already. We have

     z^2 + 25 = 0 \implies z^2 = -25 

\implies z = \pm \sqrt{-25} = \pm \sqrt{(25)(-1)} = \pm \sqrt{25}\sqrt{-1} = \pm 5\sqrt{-1} = \pm5i

    so  z = \pm5i

    So your sub of z = x+yi would give (x,y) = (0,5) and (0-5).
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    (Original post by Louis Litt)
    You've lost me already. We have

     z^2 + 25 = 0 \implies z^2 = -25 

\implies z = \pm \sqrt{-25} = \pm \sqrt{(25)(-1)} = \pm \sqrt{25}\sqrt{-1} = \pm 5\sqrt{-1} = \pm5i

    so  z = \pm5i

    So your sub of z = x+yi would give (x,y) = (0,5) and (0-5).
    great stuff, but i have firstly substituted z = x+yi into z^2 + 25 = 0
    expanded and equated real and imaginary parts, this gives me 2 equations.

    why is this not correct?
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    (Original post by YaYaD)
    great stuff, but i have firstly substituted z = x+yi into z^2 + 25 = 0
    expanded and equated real and imaginary parts, this gives me 2 equations.

    why is this not correct?
    Because  z^2 DOES NOT EQUAL  x^2 + y^2 +2iyx
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    (Original post by Louis Litt)
    Because  z^2 DOES NOT EQUAL  x^2 + y^2 +2iyx
    I know it equals x^2 + 2xyi - y^2 + 25 = 0
    when i make substitution of x+yi.

    so now why cant i equate the real and imaginary parts to 0?
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    (Original post by YaYaD)
    great stuff, but i have firstly substituted z = x+yi into z^2 + 25 = 0
    expanded and equated real and imaginary parts, this gives me 2 equations.

    why is this not correct?
    It isn't incorrect, it's justunnecessary.
    So you get out the simultaneous equations:
    x^2-y^2+25=0 and
    2xy=0
    What does this tell you? The second equation tells you that either x, y or both are 0, so now consider each possibility:
    If they're both 0 then you get 25=0, so you know that one of them must be non-zero
    Now consider y=0. Well, this is trivially false because if it were true there was no point using the substitution. Beyond that, given that x^2>0 you get the sum of two positive numbers being 0, so trivially false.
    So it must be that x=0. Now given this you just come back to your original equation but with one sign different. You're left with y^2=25 \Rightarrow y=\pm 5 \Rightarrow z=\pm 5i

    It's the very long winded route given you can just say z^2=-25 \Rightarrow z=\sqrt{-25}=\pm 5i#

    The problem you made was you forgot about the i^2 when squaring z=x+yi
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    (Original post by YaYaD)
    I know it equals x^2 + 2xyi - y^2 + 25 = 0
    when i make substitution of x+yi.

    so now why cant i equate the real and imaginary parts to 0?
    You can. It's easier than it looks: if 2xy = 0 then the only two solutions are x = 0 or y = 0. Find the value of y when x = 0 (from the other equation), and, similarly, find the value of x when y = 0. Those are the two solutions.

    I hope you can see now why the substituion of z = x+yi is pointless and silly in this context.
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    (Original post by Louis Litt)
    x and y cannot both possibly be 0 because then both simultaneous equations aren't satisfied.
    If you go on to read the rest of the post before trying to act all high and mighty then maybe you would have seen that...
    If you're going to say this then there was no point in any of that line, I could similarly say "y cannot possibly be zero because then "both" simultaneous equations aren't satisfied. I suppose while that strictly isn't a true statement, unless you treat it as true you will have an infinite number of possibilities to consider.

    Additionally, since you want to try to be a smart arse and not read I shall point out that the statement "both" aren't satisfied is also false. Last I checked 0\cdot0\cdot2=0, thus x=y=0 is a possible solution purely looking at the second equation. I would like to see you legitimately prove otherwise.
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    (Original post by Jammy Duel)
    It isn't incorrect, it's justunnecessary.
    So you get out the simultaneous equations:
    x^2-y^2+25=0 and
    2xy=0
    What does this tell you? The second equation tells you that either x, y or both are 0, so now consider each possibility:
    If they're both 0 then you get 25=0, so you know that one of them must be non-zero
    Now consider y=0. Well, this is trivially false because if it were true there was no point using the substitution. Beyond that, given that x^2>0 you get the sum of two positive numbers being 0, so trivially false.
    So it must be that x=0. Now given this you just come back to your original equation but with one sign different. You're left with y^2=25 \Rightarrow y=\pm 5 \Rightarrow z=\pm 5i

    It's the very long winded route given you can just say z^2=-25 \Rightarrow z=\sqrt{-25}=\pm 5i#

    The problem you made was you forgot about the i^2 when squaring z=x+yi
    I already multiplied the y^2 term by -1.

    But thanks i understand now.
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    (Original post by Jammy Duel)
    If you go on to read the rest of the post before trying to act all high and mighty then maybe you would have seen that...
    If you're going to say this then there was no point in any of that line, I could similarly say "y cannot possibly be zero because then "both" simultaneous equations aren't satisfied. I suppose while that strictly isn't a true statement, unless you treat it as true you will have an infinite number of possibilities to consider.

    Additionally, since you want to try to be a smart arse and not read I shall point out that the statement "both" aren't satisfied is also false. Last I checked 0\cdot0\cdot2=0, thus x=y=0 is a possible solution purely looking at the second equation. I would like to see you legitimately prove otherwise.
    Sorry I misread your post and saw my mistake so deleted my comment upon writing it.
 
 
 
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