if( a+ib)=√1+i/1-i then find the value of a^2+b^2
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ANSWER
a+ib=( 1+i1−i acha ) 100
( 1+i1−i ) 100
=( 1+i1−× 1−i1−i )
100
=(
(1)
2
−(i)
2
1−i−i+i
2
)
100
=(
1+1
1−2i−1
)
100
=(
2
−2i
)
100
=(
1
−i
)
100
=i
100
=(i
4
)
25
=1
25
{∵i
4
=1}
=1
COmparing with a+ib, we get,
a=1 and b=0
∴a+b=1
Answered By toppr
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