Question 132965
{{{((1-i)/(1+i))*((2+i)/(1+i))}}} Start with the given expression



{{{(1-i)(2+i)/(1+i)(1+i)}}} Combine the fractions



{{{(2+i-2i-i^2)/(1+i)(1+i)}}} Foil the numerator



{{{(2-i-i^2)/(1+i)(1+i)}}} Combine like terms



{{{(2-i-(-1))/(1+i)(1+i)}}} Replace {{{i^2}}} with -1



{{{(3-i)/(1+i)(1+i)}}}  Combine like terms



{{{(3-i)/(1+i+i+i^2)}}}  Foil the denominator



{{{(3-i)/(1+2i+i^2)}}}  Combine like terms



{{{(3-i)/(1+2i+(-1))}}}  Replace {{{i^2}}} with -1



{{{(3-i)/(2i)}}}  Combine like terms



{{{((3-i)/(2i))((2i)/(2i))}}}  Multiply the fraction by {{{(2i)/(2i)}}}



{{{(3-i)(2i)/(2i)(2i)}}}  Multiply the fractions



{{{(3-i)(2i)/(4i^2)}}}  {{{2i}}} and {{{2i}}} to get {{{4i^2}}} 



{{{(3-i)(2i)/(4(-1))}}}  Replace  {{{i^2}}} with -1



{{{(3-i)(2i)/(-4)}}}  Multiply



{{{(6i-2i^2)/(-4)}}}  Distribute



{{{(6i-2(-1))/(-4)}}}  Replace  {{{i^2}}} with -1



{{{(6i+2)/(-4)}}}  Multiply



{{{(2+6i)/(-4)}}}  Rearrange the terms



{{{(2)/(-4)+(6i)/(-4)}}}  Break up the fraction



{{{-1/2-(3/2)i}}}  Reduce



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Answer:



So {{{((1-i)/(1+i))*((2+i)/(1+i))}}} simplifies to {{{-1/2-(3/2)i}}} 



In other words, {{{((1-i)/(1+i))*((2+i)/(1+i))=-1/2-(3/2)i}}}