9) If
a+3i
2+ib
1-i, show that (5a-7b) = 0.
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2
Answer:
a+3i=(1-i) (2+ib)
= 2+ib-2i-i2b
=2+ib-2i-(-1)b
= 2+ib-2i+b
= (2+b)+(b-2) i
one comparing with a+3i
we get,
a=(2+b)
(b-2)=3
b=3+2=5
LHS =5a-7b
=5(2+b)-7(5)
=10+5b-35
=10+5(5)-35
=35-35
=0
RHS =0
proved
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