Respuesta :
[tex]\bf
\cfrac{5i}{2-i}\cdot \cfrac{2+i}{2+i}\impliedby \textit{multiplying by the conjugate of the bottom}\\\\
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\textit{also recall }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{and also that }i^2=-1\\\\
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[tex]\bf \cfrac{5i(2+i)}{(2-i)(2+i)}\implies \cfrac{5i(2+i)}{2^2-i^2}\implies \cfrac{5i(2+i)}{4-(-1)}\implies \cfrac{5i(2+i)}{5} \\\\\\ i(2+i)\implies 2i+i^2\implies 2i+(-1)\implies 2i-1\implies -1+2i[/tex]
[tex]\bf \cfrac{5i(2+i)}{(2-i)(2+i)}\implies \cfrac{5i(2+i)}{2^2-i^2}\implies \cfrac{5i(2+i)}{4-(-1)}\implies \cfrac{5i(2+i)}{5} \\\\\\ i(2+i)\implies 2i+i^2\implies 2i+(-1)\implies 2i-1\implies -1+2i[/tex]
