(7+4i)+(8-i)=(7+8)+(4i-i)

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Solution for (7+4i)+(8-i)=(7+8)+(4i-i) equation: 



(7+4i)+(8-i)=(7+8)+(4i-i)

We move all terms to the left:

(7+4i)+(8-i)-((7+8)+(4i-i))=0

We add all the numbers together, and all the variables

(4i+7)+(-1i+8)-(15+(+3i))=0

We get rid of parentheses

4i-1i-(15+(+3i))+7+8=0



We calculate terms in parentheses: -(15+(+3i)), so:

15+(+3i)

determiningTheFunctionDomain
(+3i)+15

We get rid of parentheses

3i+15

Back to the equation:

-(3i+15)



We add all the numbers together, and all the variables

3i-(3i+15)+15=0

We get rid of parentheses

3i-3i-15+15=0

We add all the numbers together, and all the variables

=0

i=0/1

i=0

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