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Complex number 
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#1
Apr2812, 03:42 PM

PF Gold
P: 836

The problem statement, all variables and given/known data
Given that the real and imaginary parts of the complex number [itex]z=x+iy[/itex] satisfy the equation [itex](2i)x(1+3i)y=7[/itex]. Find x and y. The attempt at a solution I know it's quite simple. Just equate the real and imaginary parts, but i checked and redid it again, but the answer still evades me! [tex](2xy7) + i(x3y)=0 \\2xy7=x \\xy=7\, (1) \\x3y=y \\4y+x=0\, (2) \\x=28/5 \\y=7/5 [/tex] I replaced in the original equation but i can't get 7 on the L.H.S. The correct answers: x=3 and y=1. 


#3
Apr2812, 04:48 PM

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P: 7,819

[itex]2xy7=x[/itex]you are saying that [itex](2xy7) + i(x3y)=z\ .[/itex]That's not what you're trying to solve ! 


#4
Apr2812, 08:27 PM

PF Gold
P: 836

Complex number
I was confused about z=x+iy. I thought i had to compare the real and imaginary parts of z with those of the equation in order to solve it. I now realize that it has absolutely nothing to do with z. All i had to do was solve the equation independently and ignore whatever was given for z.
Solving:[tex]2xy=7 \\x3y=0[/tex]I get the correct answers. Thank you, LCKurtz and SammyS. 


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