The multiplicative inverse of (1 + i) is

For z = 1 + i

Let us suppose multiplicative inverse of 1 + i is a + ib

then (1 + i) (a + ib) = 1

i.e a + ib + ai + i2b = 1

i.e a + ib + ia – b = 1

i.e $a-b+i(a+b)=1+i 0$

On comparing, real and imaginary part, we get

a – b = 1 and a + b = 0

i.e a – b  = 1 and a = – b

i.e a + a = 1

i. e $a=\frac{1}{2}$

$b=-\frac{1}{2}$

i.e multiplicative inverse of $1+i$ is $\frac{1}{2}-\frac{i}{2}$