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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 

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complex, number 
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