William Sydney Porter tried to perform the calculation [tex]\frac{-3+4i}{1+2i}[/tex]. However, he accidentally missed the minus sign, finding [tex]\frac{3+4i}{1+2i}[/tex]=[tex]\frac{11}{5}[/tex]-[tex]\frac{2i}{5}[/tex]. What answer should he have obtained?

Respuesta :

Answer:

1+2i

Step-by-step explanation:

Given the complex number [tex]\frac{-3+4i}{1+2i}[/tex] calculated by Sydney porter, In order to know the value he should have obtained if he hadn't accidentally miss the minus sign, we will have to rationalize the original complex function.

To rationalize, we will multiply the numerator and denominator if the function by the conjugate of its denominator as shown;

[tex]\frac{-3+4i}{1+2i} *\frac{1-2i}{1-2i}\\= \frac{(-3+4i)(1-2i)}{(1+2i)(1-2i)}\\= \frac{-3+6i+4i-8i^{2} }{1-2i+2i-4i^{2} } \\= \frac{-3+10i+8}{1+4} \\note\ that\ i^{2} = -1\\= \frac{5+10i}{5} \\= \frac{5}{5} + \frac{10i}{5} \\= 1 +2i[/tex]

He should have obtained 1+2i if he hadn't misses the sign