Z1=(2+3i)

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Solution for Z1=(2+3i) equation: 



1=(2+3Z)

We move all terms to the left:

1-((2+3Z))=0

We add all the numbers together, and all the variables

-((3Z+2))+1=0



We calculate terms in parentheses: -((3Z+2)), so:

(3Z+2)

We get rid of parentheses

3Z+2

Back to the equation:

-(3Z+2)



We get rid of parentheses

-3Z-2+1=0

We add all the numbers together, and all the variables

-3Z-1=0

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

-3Z=1

Z=1/-3

Z=-1/3

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