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Can someone please help me with this problem -7+4i divided by 3-2i
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(-29-2i)/13
\[\frac{-7+4i}{3-2i}=\frac{(-7+4i)(3+2i)}{(3-2i)(3+2i)}=\frac{-29-2i}{13}\]
can I ask how you got the -29 and the 13
\[(3-2i)(3+2i)=(3)^2-(2i)^2=9-4i^2=9-4(-1)=9+4=13\]
\[(-7+4i)(3+2i)=-21-14i+12i+8i^2=-21-2i-8=-29-2i\]
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hope u got it.. :)
Yes that is a great job. I get it now. Thank you very much.
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