What is the answer to z^1+z^2=(3-2i)+(4+3i) ?

The answer to z^1+z^2=(3-2i)+(4+3i) is z=2.19894…+0.18525…i,z=-3.19894…-0.18525…i

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z^1+z^2=(3-2i)+(4+3i)

z^{1} + z^{2} = (3 - 2 i) + (4 + 3 i)

z = 2.19894 \dots + 0.18525 \dots i, z = - 3.19894 \dots - 0.18525 \dots i

Solution steps

z^{1} + z^{2} = (3 - 2 i) + (4 + 3 i)

Substitute

z = x + y i

(x + y i)^{1} + (x + y i)^{2} = 3 - 2 i + 4 + 3 i

Expand

(x + x^{2} - y^{2}) + i (y + 2 x y) = 7 + i

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

[x + x^{2} - y^{2} = 7 y + 2 x y = 1]

[x + x^{2} - y^{2} = 7 y + 2 x y = 1] : (x = 2.19894 \dots, x = - 3.19894 \dots, y = 0.18525 \dots y = - 0.18525 \dots)

Substitute back

z = x + y i

z = 2.19894 \dots + 0.18525 \dots i, z = - 3.19894 \dots - 0.18525 \dots i

z^{2} + (2 i + 7) \cdot z + 6 + 3 i = 0

x, d i rec t r i xx + 5 = 0

\frac{z}{1 + z} = - 1 + i

x^{2} - 4 i x - 1 = 0

z^{2} + (2 i + 7) z + 6 + 3 i = 0

What is the answer to z^1+z^2=(3-2i)+(4+3i) ?
The answer to z^1+z^2=(3-2i)+(4+3i) is z=2.19894…+0.18525…i,z=-3.19894…-0.18525…i