Question #ea133

1 Answer
1s2s2p
Nov 11, 2017

#(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#

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