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5+5i/6+5i
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as in "divide and write in standard form?
yes
I know you multiply the conjugate of the equation in the denominator
multply top and bottom by the conjugate of the denominator the conjugate of \(a+bi\) is \(a-bi\) and this works because \[(a+bi)(a-bi)=a^2+b^2\] a real number
\[\frac{5+5i}{6+5i}\times \frac{6-5i}{6-5i}\] is step one
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the denominator is a no braniner, \[6^2+5^2=36+25=61\] all the work is in the top
43+5i/61 ?
I mean 43+15i/61
numerator is wrong
oh i see what i did wrong
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okay
55+5i/61?
yes
and don't forget two write it in standard form when you are done
try to put brackets where they should go, (55+5i)/61
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It should be in standard form so \[\frac{ 55 }{ 61 }+\frac{ 5i }{ 61 }\]
thank you
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