algebraic equations including imaginary numbers

lolserss:

so 1-( 2 over 1+i) =bi for what real number is b?

1 - 2/(1+i) = bi

1 - 2(1-i) / [(1+i)(1-i)] = bi

1 - 2(1-i) / (1- - 1) = bi

1 - 2(1-i) / 2 = bi

1 - (1-i) = bi

1-1+i = bi
i = bi
b=1