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Divide the complex numbers: (6-1) over (5+4i)
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is it 6-i? as numerator ?
I meant (6-i), yes
sry , \[6-i \div5+4i = 6-i \times5-4i/9\]
can u sove it here on ?
\[\frac{6-i}{5+4i}=\frac{6-i}{5+4i}\frac{5-4i}{5-4i}=\frac{(6-i)(5-4i)}{5^2-5\cdot4i+5\cdot4i-(4i)^2}\] \[=\frac{(6-i)(5-4i)}{25-16(-1)}=\frac{(6-i)(5-4i)}{25+16}=\frac{(6-i)(5-4i)}{41}=\cdots\]
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its right ...Zarkon.
in the end you should get \[\frac{26}{41}-\frac{29}{41}i\]
any other problem , if u have ...aria ?
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