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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