# If 2z+3-2i = (2z-i)/2 find z.

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We are given the complex equation:

2z + 3 - 2i = (2z-i) /2

We need to find z into the format of a complex number (a+bi)

First we will multiply by 2.

==> 2(2z+3-2i) = 2z-i

==> 4z +6 -4i = 2z-i

Now we will subtract 2z from both sides.

==> 2z +6 - 4i = -i

Now we will subtract 6 and add 4i to both sides.

==> 2z = -6 +4i -i

==> 2z = -6 +3i

Now we will divide by 2.

==> z = (-6/2) + (3/2)i

**==> z = -3 + (3/2)i**

2z+3-2i = (2z-i)/2 . To find z.

First we simplify the right side as z-(1/2) and then we move the z's to left and numbers to the right as below:

2z +3-2i = z- (1/2)i.

Subtract z from both sides:

2z-z + 3-2i = -(1/2)i.

z+3-2i = -(1/2)i.

Subtract 3-2i from both side:

z = -(1/2)i-(3-2i).

We add the like terms on the right.

z = -3 - 1/2i +2i.

z = -3 +(2-1/2)i.

z = -3 +(3/2)i.