What is the answer to z^2-(2i+1)/(2-i)=0 ?

The answer to z^2-(2i+1)/(2-i)=0 is z= 1/(sqrt(2))+1/(sqrt(2))i,z=-1/(sqrt(2))-1/(sqrt(2))i

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z^2-(2i+1)/(2-i)=0

z^{2} - \frac{2 i + 1}{2 - i} = 0

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

Solution steps

z^{2} - \frac{2 i + 1}{2 - i} = 0

Substitute

z = x + y i

(x + y i)^{2} - \frac{2 i + 1}{2 - i} = 0

Expand

(x + y i)^{2} - \frac{2 i + 1}{2 - i} : (x^{2} - y^{2}) + i (- 1 + 2 x y)

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

[x^{2} - y^{2} = 0 - 1 + 2 x y = 0]

[x^{2} - y^{2} = 0 - 1 + 2 x y = 0] : (x = \frac{1}{2}, x = - \frac{1}{2}, y = \frac{1}{2} y = - \frac{1}{2})

Substitute back

z = x + y i

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

∣ x + y i ∣ = y - i x

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

- y + x i = x - y i

\frac{2 + ai}{4 + bi} = 2 + 5 i

w = \frac{p + 4 i}{1 + i p}

What is the answer to z^2-(2i+1)/(2-i)=0 ?
The answer to z^2-(2i+1)/(2-i)=0 is z= 1/(sqrt(2))+1/(sqrt(2))i,z=-1/(sqrt(2))-1/(sqrt(2))i