Question

Question #ea133

Answer 1

`(2+3i)/(2i)` can be split into the partial fractions `2/(2i)+(3i)/(2i)`

By cancelling similar terms we get `(cancel(2))/(cancel(2)i)+(3cancel(i))/(2cancel(i))=1/i+3/2`

`1/i=1/sqrt(-1)`

By rationalising the denominator we get `1/sqrt(-1)*sqrt(-1)/sqrt(-1)=sqrt(-1)/sqrt(-1)^2=sqrt(-1)/-1=-sqrt(-1)=-i`

Now we have `-i+3/2=>3/2-i`