Question 949302

{{{(1+i)^8}}} in standard {{{a+bi}}}  form is:

{{{(1+i)^8=(1+i)^2*(1+i)^2*(1+i)^2*(1+i)^2}}}

since {{{(1+i)^2=1+2i+i^2=1+2i-1=2i}}}, then we have

{{{(1+i)^8=(2i)*(2i)*(2i)*(2i)}}}

since {{{(2i)*(2i)=4i^2=4(-1)=-4}}} we have

{{{(1+i)^8=(-4)*(-4)}}}

{{{(1+i)^8=16}}}