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(6-7i)/(-10+2i)
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you know what the conjugate of \(-10+2i\) is ?
is it -10-2i?
yes
multiply top and bottom by \(-10+2i\) and the reason this works is that \[(a+bi)(a-bi)=a^2+b^2\] a real number
you don't multiply by the conjugate?
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the denominator will be \[10^2+2^2=104\] the numerator will be whatever you get when you multiply \[(6-7i)(-1+2i)\]
thank you
yw typo on that last line though
yeah I noticed that, I figured
the answer key says -37+29i/52 I am not getting that and that's not what you said
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would you like help with this?
yes
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