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 March 31st, 2012, 07:22 PM #1 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 Complex numbers: Show that real part is -1 The complex number $z$ satisfies the equation $|z|=|z+2|.$Show that the real part of $z$ is $-1$. What I did was: $|z|=|z+2| let z=x+iy |x+iy|=|(x+2)+iy| \sqrt{x^2+y^2}=\sqrt{(x+2)^2+y^2} x^2+y^2=(x+2)^2+y^2 (x+2)^2-x^2=0 x^2+2x+4-x^2=0 2x=-4 x=-2$
 March 31st, 2012, 09:00 PM #2 Newbie   Joined: Apr 2010 From: kenya Posts: 28 Thanks: 0 Re: Complex numbers: Show that real part is -1 I'm afraid your attempt is only correct up to sixth step.
 March 31st, 2012, 09:47 PM #3 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 Re: Complex numbers: Show that real part is -1 Do you mean that I have expanded wrongly? I still can't spot the mistake.
 April 1st, 2012, 06:12 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Complex numbers: Show that real part is -1 You have: $(x+2)^2=x^2+2x+4$ when you should have: $(x+2)^2=x^2+4x+4$
 April 1st, 2012, 09:23 AM #5 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Complex numbers: Show that real part is -1 [color=#000000]More simple, $|z|=|z+2|\Rightarrow |z|^2=|z+2|^2\Rightarrow |z|^2=(z+2)\cdot\overline{(z+2)}\Rightarrow |z|^2=(z+2)(\overline{z}+2)\Rightarrow |z|^2=z\overline{z}+2z+2\overline{z}+4\Rightarrow$ $|z|^2=|z|^2+2(z+\overline{z})+4\Rightarrow z+\overline{z}=-2\Rightarrow 2\Re e (z)=-2\Rightarrow \Re e (z)=-1$ .[/color]

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