What is the answer to z*z+3(z-z)=4-3i ?

The answer to z*z+3(z-z)=4-3i is z= 3/(sqrt(2))-1/(sqrt(2))i,z=-3/(sqrt(2))+1/(sqrt(2))i

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z*z+3(z-z)=4-3i

z \cdot z + 3 (z - z) = 4 - 3 i

z = \frac{3}{2} - \frac{1}{2} i, z = - \frac{3}{2} + \frac{1}{2} i

Solution steps

zz + 3 (z - z) = 4 - 3 i

Substitute

z = x + y i

(x + y i) (x + y i) + 3 (x + y i - (x + y i)) = 4 - 3 i

Expand

(x + y i) (x + y i) + 3 (x + y i - (x + y i)) : (x^{2} - y^{2}) + 2 i x y

(x^{2} - y^{2}) + 2 i x y = 4 - 3 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} - y^{2} = 4 2 x y = - 3]

[x^{2} - y^{2} = 4 2 x y = - 3] : (x = \frac{3}{2}, x = - \frac{3}{2}, y = - \frac{1}{2} y = \frac{1}{2})

Substitute back

z = x + y i

z = \frac{3}{2} - \frac{1}{2} i, z = - \frac{3}{2} + \frac{1}{2} i

x + y i = (3 + i) (2 - 3 i)

z^{2} - 4 i z - 6 = 0

z^{2} - (3 - 2 i) z + (5 - i) = 0

x + y i = (1 - i) (2 + 8 i)

2 (a + ib) + (3 + i) (a - ib) = 4 - i

What is the answer to z*z+3(z-z)=4-3i ?
The answer to z*z+3(z-z)=4-3i is z= 3/(sqrt(2))-1/(sqrt(2))i,z=-3/(sqrt(2))+1/(sqrt(2))i