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FP1 Complex Numbers

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Reply 20

brianeverit

Original post by krisshP

But my textbook says that
z=2+3i, w=1-2i
z=2+I, w=3-2i

Is the book wrong?

I apologise for suggesting a method which only works if z and w are real numbers not complex.

Reply 21

davros

Original post by brianeverit

I apologise for suggesting a method which only works if z and w are real numbers not complex.

Don't worry - I skim-read the 1st page and thought your method was going to work out really neat! Then when I came back to the question I wrote down 4-4i instead of 4+4i and ended up with a horrendous quadratic with awkward roots.

As I've pointed out to the OP, if you just go with the substitution method like a normal set of equations, the quadratic in w comes out quite nicely :smile:

Reply 22

krisshP

Original post by davros

Book's right I'm afraid!

I did a load of horrendous working out earlier based on the wrong equation and thought the book was wrong, but actually you can solve the quadratic derived by Khalil in the usual way (i.e. the quadratic formula) to get. the solutions you need.

You should get to

w^2 + 4(i-1)w - 8i - 1 = 0

- plug this mess into the quadratic formula and the discriminant works out quite nicely :smile:

b^2 - 4ac= 16i^2 - 32i + 16 + 32i + 4
=4
w=-2i + 2 ±1
So w=3-2i or w=1-2i (just like book answers)

The rest should be w piece of cake :smile:

Finally this question is cracked!!!!!!!: D

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