(3+4i)/(2-i)+(3+4i)/(2+i)=0

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Solution for (3+4i)/(2-i)+(3+4i)/(2+i)=0 equation: 



(3+4i)/(2-i)+(3+4i)/(2+i)=0



Domain of the equation: (2-i)!=0

We move all terms containing i to the left, all other terms to the right

-i!=-2

i!=-2/-1

i!=+2

i∈R





Domain of the equation: (2+i)!=0

We move all terms containing i to the left, all other terms to the right

i!=-2

i∈R



We add all the numbers together, and all the variables

(4i+3)/(-1i+2)+(4i+3)/(i+2)=0

We calculate fractions


((4i+3)*(i+2))/((-1i+2)*(i+2))+((4i+3)*(-1i+2))/((-1i+2)*(i+2))=0



We calculate terms in parentheses: +((4i+3)*(i+2))/((-1i+2)*(i+2)), so:

(4i+3)*(i+2))/((-1i+2)*(i+2)

We multiply all the terms by the denominator


(4i+3)*(i+2))

Back to the equation:

+((4i+3)*(i+2)))





We calculate terms in parentheses: +((4i+3)*(-1i+2))/((-1i+2)*(i+2)), so:

(4i+3)*(-1i+2))/((-1i+2)*(i+2)

We multiply all the terms by the denominator


(4i+3)*(-1i+2))

Back to the equation:

+((4i+3)*(-1i+2)))





We calculate terms in parentheses: +((4i+3)*(i+2)))+((4i+3)*(-1i+2))), so:

(4i+3)*(i+2)))+((4i+3)*(-1i+2))

determiningTheFunctionDomain
(4i+3)*(i+2)))+((4i-1i+3)*(+2))

We add all the numbers together, and all the variables

(4i+3)*(i+2)))+((3i+3)*2)

We add all the numbers together, and all the variables

(4i+3)*(i+2)))+((3i

Back to the equation:

+((4i+3)*(i+2)))+((3i)

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