Simple complex equation

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

$z^2 + 25 = 0 \implies z^2 = -25 [br]\implies z = \pm \sqrt{-25} = \pm \sqrt{(25)(-1)} = \pm \sqrt{25}\sqrt{-1} = \pm 5\sqrt{-1} = \pm5i$

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?

Because $z^2$ DOES NOT EQUAL $x^2 + y^2 +2iyx$

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?

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?

x and y cannot both possibly be 0 because then both simultaneous equations aren't satisfied.

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

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.