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 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 24th, 2015, 04:32 PM #11 Newbie   Joined: Jul 2015 From: Serbia Posts: 8 Thanks: 1 $\begin{array}{l} \left( {4 + 3i} \right)z = 2 - i\\ z = \frac{{2 - i}}{{4 + 3i}}\frac{{4 - 3i}}{{4 - 3i}} = \frac{{8 - 6i - 4i + 3{i^2}}}{{{4^2} + {3^2}}} = \frac{{5 - 10i}}{{25}} = \frac{{5\left( {1 - 2i} \right)}}{{25}} = \frac{1}{5}\left( {1 - 2i} \right) \end{array}$ July 24th, 2015, 04:40 PM #12 Newbie   Joined: Jul 2015 From: Serbia Posts: 8 Thanks: 1 $\begin{array}{l} \left( {4 + 3i} \right)z = 2 - i\\ z = x + iy\\ \left( {4 + 3i} \right)\left( {x + iy} \right) = 2 - i\\ 4x + i4y + i3x - 3y = 2 - i\\ 4x - 3y + i\left( {3x + 4y} \right) = 2 - i\\ 4x - 3y = 2/ \cdot 4\\ 3x + 4y = - 1/ \cdot 3\\ 16x - 12y = 8\\ 9x + 12y = - 3\\ 25x = 5\\ x = \frac{1}{5}\\ 3y = 4x - 2 = 4 \cdot \frac{1}{5} - 2 = \frac{4}{5} - \frac{{10}}{5} = - \frac{6}{5}\\ y = \frac{{ - \frac{6}{5}}}{3} = - \frac{2}{5}\\ z = x + iy = \frac{1}{5} - i\frac{2}{5} \end{array}$ July 28th, 2015, 09:05 PM #13 Newbie   Joined: Jul 2015 From: jalandhar Posts: 1 Thanks: 0 Math Focus: algebra i can not help you because am weak in calculations  Tags complex, number 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 1729 Algebra 17 January 31st, 2015 05:00 AM mhhojati Algebra 7 July 28th, 2012 10:03 AM Punch Complex Analysis 1 April 1st, 2012 09:01 AM panky Algebra 4 November 14th, 2011 10:15 PM TsAmE Complex Analysis 1 October 18th, 2010 04:38 PM

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