Which values of p and q would make the value of the following expression equal to 29i? (2 – 5i)(p + q)i

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Which values of p and q would make the value of the following expression equal to 29i? (2 – 5i)(p + q)i

Mathematics
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Ok, expand the (2-5i)(p+q)i and what do you then get?
expand?
yes in other words multiply out the bracket

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its still confusing
\(\large\color{black}{ \displaystyle (a+b)(c+d)=ab+ac+bc+bd }\) have you seen anything like this before?
yes
i need to look into which property is which still get them confused
\(\large\color{black}{ \displaystyle (2 – 5i)(p + q)=2(p+q)-5i(p+q) }\) same way as: \((a-b)(C)=a(C)-b(C)\) ---------------------------------------- And therefore: \(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)-5i^2(p+q) }\) we know \(i^2=-1\) because \(i^2=\left(\sqrt{-1}\right)^2=-1,\) like any \(\left(\sqrt{a}\right)^2=a\) So, \(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)-5(-1)(p+q) }\) \(\large\color{black}{ \displaystyle (2 – 5i)(p + q)i=2i(p+q)+5(p+q) }\)
Now, how would you get 29i in a "a+bi" form? 0+29i and so here we have this form as well 5(p+q) + 2(p+q)i so a=0=5(p+q) bi=29i=2(p+q)i
solve the system 0=5(p+q) 29=2(p+q) (---> this eq. results after canceling the i's on both sides.)
good luck

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