Answer
$-1-i$
Work Step by Step
Multiplying both the numerator and the denominator of $
\dfrac{-2i}{1+i}
$ by the complex conjugate of the denominator, then
\begin{align*}\require{cancel}
&
\dfrac{-2i}{1+i}\cdot\dfrac{1-i}{1-i}
\\\\&=
\dfrac{-2i(1)-2i(-i)}{(1+i)(1-i)}
&\text{ (use Distributive Property)}
\\\\&=
\dfrac{-2i(1)-2i(-i)}{(1)^2-(i)^2}
&\text{ (use $(a+b)(a-b)=a^2-b^2$)}
\\\\&=
\dfrac{-2i+2i^2}{1-i^2}
\\\\&=
\dfrac{-2i+2(-1)}{1-(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{-2i-2}{1+1}
\\\\&=
\dfrac{-2i-2}{2}
\\\\&=
\dfrac{-\cancel2^1i-\cancel2^1}{\cancel2^1}
&\text{ (divide by $2$)}
\\\\&=
-i-1
\\\\&=
-1-i
.\end{align*}
Hence, the simplified form of the given expression is $
-1-i
$.
