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

The answer to z^2-z^1=(4+3i)-(3-2i) is z=2.28939…+1.39711…i,z=-1.28939…-1.39711…i

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

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

z = 2.28939 \dots + 1.39711 \dots i, z = - 1.28939 \dots - 1.39711 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

(x^{2} - y^{2} - x) + i (- y + 2 x y) = 1 + 5 i

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

[x^{2} - y^{2} - x = 1 - y + 2 x y = 5]

[x^{2} - y^{2} - x = 1 - y + 2 x y = 5] : (x = 2.28939 \dots, x = - 1.28939 \dots, y = 1.39711 \dots y = - 1.39711 \dots)

Substitute back

z = x + y i

z = 2.28939 \dots + 1.39711 \dots i, z = - 1.28939 \dots - 1.39711 \dots i

5 x + 7 y i = - x - 7 i

z^{2} - (1 - 2 i) z + (2 + 14 i) = 0

3 b - 2 ai = 4 - 3 i

∣ z + 3 - 4 i ∣ = 2

2 x + 9 y i = 16 - 12 i

What is the answer to z^2-z^1=(4+3i)-(3-2i) ?
The answer to z^2-z^1=(4+3i)-(3-2i) is z=2.28939…+1.39711…i,z=-1.28939…-1.39711…i