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[tex](2-5i)(p+q)i=29i \ \ \ \ \ \ \ \ |\div i \\ (2-5i)(p+q)=29 \ \ \ \ \ \ \ \ \ \ |\div (2-5i) \\ p+q=\frac{29}{2-5i} \\ \\ \overline{\underline{\frac{29}{2-5i}=\frac{29(2+5i)}{(2-5i)(2+5i)}=\frac{29(2+5i)}{2^2-(5i)^2}=\frac{29(2+5i)}{4-25i^2}=\frac{29(2+5i)}{4+25}=\frac{29(2+5i)}{29}=2+5i}} \\ \\ p+q=2+5i[/tex]

In your answer options p is a real number and q is an imaginary number, so p=2 and q=5i.

The answer is A.

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