What is the answer to (z-3i)+z+4i^3=8i(3i+3i^2) ?

The answer to (z-3i)+z+4i^3=8i(3i+3i^2) is z=-12-i 17/2

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(z-3i)+z+4i^3=8i(3i+3i^2)

(z - 3 i) + z + 4 i^{3} = 8 i (3 i + 3 i^{2})

z = - 12 - i \frac{17}{2}

Solution steps

(z - 3 i) + z + 4 i^{3} = 8 i (3 i + 3 i^{2})

Simplify

(z - 3 i) + z + 4 i^{3} : 2 z - 7 i

Simplify

8 i (3 i + 3 i^{2}) : - 24 - 24 i

2 z - 7 i = - 24 - 24 i

Move

7 i

to the right side

2 z = - 24 - 17 i

Divide both sides by

2

z = - 12 - i \frac{17}{2}

x, 60 x + 120 y = 500

y = 3 p^{2} - 4

lo g_{10} (0.9) = x

x, 7 x - 3 + 4 y = 27

z = \frac{( 4 + 5 i )}{( 7 + 2 i )}

What is the answer to (z-3i)+z+4i^3=8i(3i+3i^2) ?
The answer to (z-3i)+z+4i^3=8i(3i+3i^2) is z=-12-i 17/2