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# FP1 simultaneous eq watch

1. Really stuck on the following question (Q15 FP1 p64)
Given that λ ∈ ℝ and that z and w are complex numbers, solve the simultaneous equations z - iw = 2, z - λw = 1 - λ^2, giving your answers in the form a + ib, where a, b ∈ ℝ, and a and b are functions of λ.
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I got up to:

w = (i(-1-λ^2) + λ(-1-λ^2)) / -1-λ^2
w = i + λ

--

the answer is 1 + iλ

where've I gone wrong!?

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For the next question, I'm given that 5-2i is a root of z^2 - 10z + c =0. How to approach this? I know that z - 5 + 2i will be a factor.
2. For your first one, you haven't finished solving it completely yet. You have just got what 'w' is, you still need to find 'z'.
3. (Original post by claret_n_blue)
For your first one, you haven't finished solving it completely yet. You have just got what 'w' is, you still need to find 'z'.
I just facepalmed... have you any idea about the second?
4. (Original post by Lunch_Box)
I just facepalmed... have you any idea about the second?

What I would recommend is this. Forget about the fact that the quadratic function is in terms of 'z', think of it as any normal quadratic, how would you solve this? Seeing as you can't see obvious roots here, you would use the quadratic formula, right?

So now you get the formula in terms of 'c'. Also, you know that one of your roots (solutions) to the formula is z = 5 - 2i. Get what to do from here?
5. (Original post by claret_n_blue)

What I would recommend is this. Forget about the fact that the quadratic function is in terms of 'z', think of it as any normal quadratic, how would you solve this? Seeing as you can't see obvious roots here, you would use the quadratic formula, right?

So now you get the formula in terms of 'c'. Also, you know that one of your roots (solutions) to the formula is z = 5 - 2i. Get what to do from here?
I got up to:

5 - 2i = 10 - √c (gives answer as a+bi)
5- 2i = -√c (gives answer as a+bi)

can't seem to get c = 29
6. (Original post by Lunch_Box)
I got up to:

5 - 2i = 10 - √c (gives answer as a+bi)
5- 2i = -√c (gives answer as a+bi)

can't seem to get c = 29
It shouldn't be , that bit is wrong. How did you get to this bit?
7. (Original post by claret_n_blue)
It shouldn't be , that bit is wrong. How did you get to this bit?
x = (10 + 10 - 2√c) /2
8. For the next question, I'm given that 5-2i is a root
of z^2 - 10z + c =0. How to approach this? I
know that z - 5 + 2i will be a factor.

You have two complex roots and you know that they are solutions to an equation. Put them in the form (x-a)(x-b) and expand.

(x-(5-2i))(x-(5+2i))
9. (Original post by claret_n_blue)

Seeing as you can't see obvious roots here, you would use the quadratic formula, right?
is a root, then so is . Done.
10. Above post. Z=X
11. brb killing myself
12. (Original post by Lunch_Box)
x = (10 + 10 - 2√c) /2
No. Come on, think. You know the quadratic equation:

You correctly identified and so I would expect there to be a somewhere in there.

Don't try and skip loads of steps, do it step by step and do it properly.
is a root, then so is . Done.
Good point.

OP this is another way of doing it as well. With your original root give (z = 5 - 2i), you know what your complex conjugate is. Multiplying these two together will give you your quadratic (if you've done it correctly) and then you can simply see what 'c' is.

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