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Sagot :
Given the below
[tex](\frac{5}{2}+2i)(\frac{1}{4}-6i)[/tex]Multiplication of the vectors in the a+bi form gives
Applying the complex arithmetic rule below
[tex](a+bi)(c+di)=(ac-bd)(ad+bc)i[/tex]The expansion of the vectors gives
[tex](\frac{5}{2}+2i)(\frac{1}{4}-6i)=\frac{5}{2}(\frac{1}{4}-6i)+2i(\frac{1}{4}-6i)[/tex]Opening the brackets
[tex]\begin{gathered} (\frac{5}{2}+2i)(\frac{1}{4}-6i)=\frac{5}{2}(\frac{1}{4}-6i)+2i(\frac{1}{4}-6i) \\ =\frac{5}{8}-\frac{30}{2}i+\frac{2}{4}i-12i^2 \\ \text{Where i}^2=-1 \\ =\frac{5}{8}-\frac{30}{2}i+\frac{2}{4}i-12(-1) \\ ==\frac{5}{8}+12-\frac{30}{2}i+\frac{2}{4}i \end{gathered}[/tex]Simplifying the above expression
[tex]\begin{gathered} =\frac{5+96}{8}+\frac{-60i+2i}{4}=\frac{101}{8}+\frac{-58}{4}i \\ =\frac{101}{8}+\frac{-29}{2}i=\frac{101}{8}-\frac{29i}{2} \\ (\frac{5}{2}+2i)(\frac{1}{4}-6i)=\frac{101}{8}-\frac{29i}{2} \end{gathered}[/tex]Hence, the answer is
[tex]=\frac{101}{8}-\frac{29i}{2}[/tex]
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