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 November 2nd, 2017, 03:00 PM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Solve complex number equations with conjugate I have this equation $\displaystyle \displaystyle z + 2\bar{z} = 2 - i$, but don't really know how to solve it. My first thought was to set $\displaystyle \displaystyle z = \bar{z}$ and solve for $\displaystyle z$. And when z was figured out I put in z and $\displaystyle \displaystyle \bar{z}$, but that failed. So how do you solve these types of equations? November 2nd, 2017, 03:30 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,575 Thanks: 1423 let $z=x+i y,~x,y \in \mathbb{R}$ $x + i y + 2x - 2i y = 2 - i$ $3x - i y = 2 - i$ $3x = 2$ $-y = -1$ $x=\dfrac 2 3$ $y = 1$ $z = \dfrac 2 3 + i$ Thanks from DecoratorFawn82 November 2nd, 2017, 05:28 PM #3 Senior Member   Joined: Aug 2012 Posts: 2,409 Thanks: 753 Here's another way using just using the standard operations on the complex numbers: conjugation, Real part $\Re$ and Imaginary part $\Im$. (1) $z + 2 \bar z = 2− i$ $\ \ \ \$ [Given] (2) $\bar z + 2 z = 2 + i$ $\ \ \ \$ [Conjugating both sides] $3 z + 3 \bar z = 4$ $\ \ \ \$ [Adding (1) to (2)] $z + \bar z = \frac{4}{3}$ $2 \Re (z) = \frac{4}{3}$ $\ \ \ \$ [Using $z + \bar z = 2 \Re (z)$] $\Re (z) = \frac{2}{3}$ Now subtracting (2) - (1) gives: $z - \bar z = 2$ and then $\Im (z) = 1$ $\ \ \ \$ [Using $z - \bar z = 2 \Im (z)$] So $z = \frac{2}{3} + i$ Thanks from topsquark, DecoratorFawn82 and romsek Last edited by Maschke; November 2nd, 2017 at 05:31 PM. November 2nd, 2017, 08:39 PM   #4
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 Originally Posted by Maschke $\Re$ and $\Im$.
People still use that ugly script for those? I haven't seen that notation in years!

-Dan November 2nd, 2017, 09:37 PM   #5
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 Originally Posted by topsquark People still use that ugly script for those? I haven't seen that notation in years! -Dan
I hate them too. That's how \Re and \Im render. Perhaps I should just spell them out as Re and Im. $Re(z)$ and $Im(z)$. Yes that's much better. I'll do that from now on. November 2nd, 2017, 10:05 PM   #6
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 Originally Posted by Maschke I hate them too. That's how \Re and \Im render. Perhaps I should just spell them out as Re and Im. $Re(z)$ and $Im(z)$. Yes that's much better. I'll do that from now on.
My tensor geometry book has a comment about the labeling of tensors, which uses symbols very similar to the ones LaTeX wants you to use. The author claimed that the notation was ugly and did nothing to help the reader. So he said he wasn't going to use that script. Looking down at the footnote where the author said this he got a "thanks" from his typist. -Dan November 2nd, 2017, 10:06 PM   #7
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 Originally Posted by Maschke Re and Im. $Re(z)$ and $Im(z)$.
That's what us Physics people use.

-Dan November 22nd, 2017, 05:43 PM #8 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 673 Thanks: 88 I haven't seen a z with a bar on top before. What does it mean? November 22nd, 2017, 06:16 PM   #9
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 Originally Posted by EvanJ I haven't seen a z with a bar on top before. What does it mean?
Complex conjugate. If $z = a + bi$ then $\bar z = a - bi$. It's $z$ reflected in the real axis. Tags complex, conjugate, equations, number, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ehh Pre-Calculus 1 January 17th, 2015 08:15 PM bilano99 Algebra 4 February 13th, 2013 01:20 PM gab633 Linear Algebra 1 September 30th, 2011 03:55 AM ecogreen Algebra 1 June 13th, 2010 05:11 AM bilano99 Calculus 1 December 31st, 1969 04:00 PM

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