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July 24th, 2015, 04:32 PM   #11
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\[\begin{array}{l}
\left( {4 + 3i} \right)z = 2 - i\\
z = \frac{{2 - i}}{{4 + 3i}}\frac{{4 - 3i}}{{4 - 3i}} = \frac{{8 - 6i - 4i + 3{i^2}}}{{{4^2} + {3^2}}} = \frac{{5 - 10i}}{{25}} = \frac{{5\left( {1 - 2i} \right)}}{{25}} = \frac{1}{5}\left( {1 - 2i} \right)
\end{array}\]
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July 24th, 2015, 04:40 PM   #12
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\[\begin{array}{l}
\left( {4 + 3i} \right)z = 2 - i\\
z = x + iy\\
\left( {4 + 3i} \right)\left( {x + iy} \right) = 2 - i\\
4x + i4y + i3x - 3y = 2 - i\\
4x - 3y + i\left( {3x + 4y} \right) = 2 - i\\
4x - 3y = 2/ \cdot 4\\
3x + 4y = - 1/ \cdot 3\\
16x - 12y = 8\\
9x + 12y = - 3\\
25x = 5\\
x = \frac{1}{5}\\
3y = 4x - 2 = 4 \cdot \frac{1}{5} - 2 = \frac{4}{5} - \frac{{10}}{5} = - \frac{6}{5}\\
y = \frac{{ - \frac{6}{5}}}{3} = - \frac{2}{5}\\
z = x + iy = \frac{1}{5} - i\frac{2}{5}
\end{array}\]
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July 28th, 2015, 09:05 PM   #13
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Math Focus: algebra
i can not help you because am weak in calculations
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