InterviewSolution
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InterviewSolution
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Reduce `((1)/(1+2i)+(3)/(1-i))((3-2i)/(1+3i))` to the form `(a + ib).` |
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Answer» We have `((1)/(1+2i)+(3)/(1-i))((3-2i)/(1+3i))` `={((1-i)+(3+6i))/((1+2i)(1-i))}((3-2i)/(1+3i))=((4+5i))/((3+i))xx((3-2i))/((1+3i))` `=((4+5i)(3-2i))/((3+i)(1+3i))=((12+10)+(15-8)i)/((3-3)+(9i+i))=((22+7i))/(10i)xx(i)/(i)` `=((7i^(2)+22i))/(10i^(2))=(-7+22i)/(-10)=((7)/(10)-(22)/(10)i)=((7)/(10)-(11)/(5)i)`. |
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