multiplicative inverse of (1+i)^2/(3-i)
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=(1+i)^2/3-i
=1-1+2i/3-i *3+i/3+i
=2i*(3+i)/3^2-i^2
=6i-2/9+1
=2(3i-1)/10
=3i-1/5
multiplicative inverse = 5/3i-1
=1-1+2i/3-i *3+i/3+i
=2i*(3+i)/3^2-i^2
=6i-2/9+1
=2(3i-1)/10
=3i-1/5
multiplicative inverse = 5/3i-1
sanjaivnair2000:
it should be again multiplied by the conjugate right?
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