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

The answer to (i*z-3)/(z^2-1)=-1 is z=(sqrt(15))/2-1/2 i,z=-(sqrt(15))/2-1/2 i

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(i*z-3)/(z^2-1)=-1

\frac{i \cdot z - 3}{z ^{2} - 1} = - 1

z = \frac{15}{2} - \frac{1}{2} i, z = - \frac{15}{2} - \frac{1}{2} i

Solution steps

\frac{i z - 3}{z ^{2} - 1} = - 1

Substitute

z = x + y i

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

Multiply both sides by

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

\frac{i ( x + y i ) - 3}{( x + y i ) ^{2} - 1} ((x + y i)^{2} - 1) = - 1 \cdot ((x + y i)^{2} - 1)

Expand

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

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

[- y - 3 = - x^{2} + y^{2} + 1 x = - 2 x y]

[- y - 3 = - x^{2} + y^{2} + 1 x = - 2 x y] : (x = \frac{15}{2}, x = - \frac{15}{2}, y = - \frac{1}{2} y = - \frac{1}{2})

Substitute back

z = x + y i

z = \frac{15}{2} - \frac{1}{2} i, z = - \frac{15}{2} - \frac{1}{2} i

4 i = x^{2} - 3 x + 7

(1 + i 3)^{11} = a + ib

a + bi + 0 + 0 i = a + bi

x, f i = \frac{g r}{x}

z^{2} - (5 - 4 i) z + 20 i = 0

What is the answer to (i*z-3)/(z^2-1)=-1 ?
The answer to (i*z-3)/(z^2-1)=-1 is z=(sqrt(15))/2-1/2 i,z=-(sqrt(15))/2-1/2 i