complex numbers

RauLiTo · 2006-10-31 04:22:53

get X and Y in the below equation :
X (X+iY)² - 10(X+iY) + 14 = 0

help please guys i am getting crazy

RauLiTo · 2006-10-31 05:21:29

guys ?

luca-deltodesco · 2006-10-31 05:33:18

if that first x wasnt there, you could use quadratic equation, but since it is, i dont know

RauLiTo · 2006-10-31 06:35:12

:d anybody else ?

RauLiTo · 2006-10-31 07:46:03

come on guys !!! is my question is not clear or what ?

tom112 · 2006-11-08 03:03:40

First, expand the brackets...
x(x + iy)^2 becomes x(x^2 + 2xyi - y^2)
which in turn becomes x^3 + 2(x^2)yi - xy^2

the second term expands to -10x - 10iy

So the equation with the brackets expanded reads

x^3 + 2(x^2)yi - xy^2 -10x - 10iy + 14 = 0

Simplify like terms and generally tidy things up:

x^3 -xy^2 - 10x + 14 + i(2(x^2)y - 10y) = 0 (+ 0i), this makes the next step slightly clearer...

Equate real and imag. coeffs:
x^3 -xy^2 - 10x + 14 = 0
=> y = (x^2 - 1 - 10 + 14)^1/2
=> y = (x^2 + 3)^1/2 (Not very pretty...)

2(x^2)y - 10y = 0
=> x^2 - 5 = 0
=> x = + or - 5^1/2
The + or - bit becomes irrelevant, since the x is squared in the previous equation.
=> y = (5 + 3)^1/2 = sqrt(8) (supposed to be a 9?)

So your solutions (possibly) are x = +/- sqrt(5), y = sqrt(8)
I highly doubt I have gone straight through this without making some sort of error, so please check it - the idea was to give you and idea of which process to undertake when solving this questions.

Math Is Fun Forum

#1 2006-10-31 04:22:53

complex numbers

#2 2006-10-31 05:21:29

Re: complex numbers

#3 2006-10-31 05:33:18

Re: complex numbers

#4 2006-10-31 06:35:12

Re: complex numbers

#5 2006-10-31 07:46:03

Re: complex numbers

#6 2006-11-08 03:03:40

Re: complex numbers

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