Respuesta :
Answer:
[tex]\frac{12}{13}+\frac{18}{13}i[/tex]
Step-by-step explanation:
Multiply by the conjugate as a factor of one:
[tex]\frac{6i}{3+2i}=\frac{6i}{3+2i}*\frac{3-2i}{3-2i}=\frac{18i+12}{9+4}=\frac{12+18i}{13}=\frac{12}{13}+\frac{18}{13}i[/tex]
Answer: [tex]\frac{12}{13}+\frac{18}{13}i[/tex]
The expression is in a+bi form where,
a = 12/13
b = 18/13
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Work Shown:
Multiply top and bottom by 3-2i, which is the conjugate of the original denominator. This will make the denominator go from a nonreal complex number to a purely real number.
[tex]\frac{6i}{3+2i}\\\\\\\frac{6i(3-2i)}{(3+2i)(3-2i)}\\\\\\\frac{18i-12i^2}{(3)^2-(2i)^2}\\\\\\\frac{18i-12i^2}{9-4i^2}\\\\\\\frac{18i-12(-1)}{9-4(-1)}\\\\\\\frac{18i+12}{9+4}\\\\\\\frac{12+18i}{13}\\\\\\\frac{12}{13}+\frac{18}{13}i\\\\\\[/tex]
The last expression is in a+bi form where a = 12/13 and b = 18/13
Keep in mind that [tex]i = \sqrt{-1}[/tex] has both sides square to [tex]i^2 = -1[/tex]. Also, I'm using the difference of squares rule to go from [tex](3+2i)(3-2i)[/tex] to [tex](3)^2 - (2i)^2[/tex].
