What is the answer to (z^2-i)/(z-2)=sqrt(2) ?

The answer to (z^2-i)/(z-2)=sqrt(2) is z=0.38643…-1.55924…i,z=1.02777…+1.55924…i

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(z^2-i)/(z-2)=sqrt(2)

\frac{z ^{2} - i}{z - 2} = 2

z = 0.38643 \dots - 1.55924 \dots i, z = 1.02777 \dots + 1.55924 \dots i

Solution steps

\frac{z ^{2} - i}{z - 2} = 2

Substitute

z = x + y i

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

Multiply both sides by

x - 2 + i y

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

Expand

(x^{2} - y^{2}) + i (- 1 + 2 x y) = (2 x - 22) + 2 i y

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

[x^{2} - y^{2} = 2 x - 22 - 1 + 2 x y = 2 y]

[x^{2} - y^{2} = 2 x - 22 - 1 + 2 x y = 2 y] : (x = 0.38643 \dots, x = 1.02777 \dots, y = - 1.55924 \dots y = 1.55924 \dots)

Substitute back

z = x + y i

z = 0.38643 \dots - 1.55924 \dots i, z = 1.02777 \dots + 1.55924 \dots i

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

(1 + i) x + (- 2 + i) y = - 3 + 16 i

24 + 42 y i = - 0.5 x + 0.5 i

6 (4 + 7 i)^{2} + A (4 + 7 i) + B = 0

x + i y = 1, 1 x + y = 1 + i

What is the answer to (z^2-i)/(z-2)=sqrt(2) ?
The answer to (z^2-i)/(z-2)=sqrt(2) is z=0.38643…-1.55924…i,z=1.02777…+1.55924…i