What is ((z^2+2*z+I))/((1-z))=-1 ?

The solution to ((z^2+2*z+I))/((1-z))=-1 is Parabola with (h,k)=(-1/2 ,-3/4),p=-1/4

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((z^2+2*z+I))/((1-z))=-1

\frac{( z ^{2} + 2 \cdot z + I )}{( 1 - z )} = - 1

(h, k) = (- \frac{1}{2}, - \frac{3}{4}), p = - \frac{1}{4}

Solution steps

\frac{z ^{2} + 2 z + I}{1 - z} = - 1

Rewrite

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

in the standard form

4 (- \frac{1}{4}) (I - (- \frac{3}{4})) = (z - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, - \frac{3}{4}), p = - \frac{1}{4}

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

(x + 1)^{2} = - 8 (y - 4)

x = - 7 y^{2}

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

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

What is ((z^2+2*z+I))/((1-z))=-1 ?
The solution to ((z^2+2*z+I))/((1-z))=-1 is Parabola with (h,k)=(-1/2 ,-3/4),p=-1/4