33+i=3(1-3i)

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Solution for 33+i=3(1-3i) equation: 



33+i=3(1-3i)

We move all terms to the left:

33+i-(3(1-3i))=0

We add all the numbers together, and all the variables

i-(3(-3i+1))+33=0



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

3(-3i+1)

We multiply parentheses

-9i+3

Back to the equation:

-(-9i+3)



We get rid of parentheses

i+9i-3+33=0

We add all the numbers together, and all the variables

10i+30=0

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

10i=-30

i=-30/10

i=-3

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