Solve 2z=1-i/1+z+2i

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

Solve for z

$$z=\frac{\sqrt{2}}{2}+\left(-\frac{1}{2}+\frac{1}{2}i\right)\approx 0.207106781+0.5i$$

$$z=-\frac{1}{2}+\frac{1}{2}i-\frac{\sqrt{2}}{2}\approx -1.207106781+0.5i$$

Steps Using the Quadratic Formula

Variable z cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by z+1.

$$2z\left(z+1\right)=1-i+\left(z+1\right)\times \left(2i\right)$$

Use the distributive property to multiply 2z by z+1.

$$2z^{2}+2z=1-i+\left(z+1\right)\times \left(2i\right)$$

Use the distributive property to multiply z+1 by 2i.

$$2z^{2}+2z=1-i+2iz+2i$$

Do the additions in 1-i+2i.

$$2z^{2}+2z=2iz+1+i$$

Subtract 2iz from both sides.

$$2z^{2}+2z-2iz=1+i$$

Combine 2z and -2iz to get \left(2-2i\right)z.

$$2z^{2}+\left(2-2i\right)z=1+i$$

Subtract 1+i from both sides.

$$2z^{2}+\left(2-2i\right)z-\left(1+i\right)=0$$

Multiply -1 and 1+i to get -1-i.

$$2z^{2}+\left(2-2i\right)z+\left(-1-i\right)=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2-2i for b, and -1-i for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-2+2i±\sqrt{\left(2-2i\right)^{2}-4\times 2\left(-1-i\right)}}{2\times 2}$$

Square 2-2i.

$$z=\frac{-2+2i±\sqrt{-8i-4\times 2\left(-1-i\right)}}{2\times 2}$$

Multiply -4 times 2.

$$z=\frac{-2+2i±\sqrt{-8i-8\left(-1-i\right)}}{2\times 2}$$

Multiply -8 times -1-i.

$$z=\frac{-2+2i±\sqrt{-8i+\left(8+8i\right)}}{2\times 2}$$

Add -8i to 8+8i.

$$z=\frac{-2+2i±\sqrt{8}}{2\times 2}$$

Take the square root of 8.

$$z=\frac{-2+2i±2\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$z=\frac{-2+2i±2\sqrt{2}}{4}$$

Now solve the equation z=\frac{-2+2i±2\sqrt{2}}{4} when ± is plus. Add -2+2i to 2\sqrt{2}.

$$z=\frac{-2+2i+2\sqrt{2}}{4}$$

Divide -2+2i+2\sqrt{2} by 4.

$$z=\frac{\sqrt{2}}{2}+\left(-\frac{1}{2}+\frac{1}{2}i\right)$$

Now solve the equation z=\frac{-2+2i±2\sqrt{2}}{4} when ± is minus. Subtract 2\sqrt{2} from -2+2i.

$$z=\frac{-2+2i-2\sqrt{2}}{4}$$

Divide -2+2i-2\sqrt{2} by 4.

$$z=-\frac{1}{2}+\frac{1}{2}i-\frac{\sqrt{2}}{2}$$

The equation is now solved.

$$z=\frac{\sqrt{2}}{2}+\left(-\frac{1}{2}+\frac{1}{2}i\right) z=-\frac{1}{2}+\frac{1}{2}i-\frac{\sqrt{2}}{2}$$

Steps for Completing the Square

Variable z cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by z+1.

$$2z\left(z+1\right)=1-i+\left(z+1\right)\times \left(2i\right)$$

Use the distributive property to multiply 2z by z+1.

$$2z^{2}+2z=1-i+\left(z+1\right)\times \left(2i\right)$$

Use the distributive property to multiply z+1 by 2i.

$$2z^{2}+2z=1-i+2iz+2i$$

Do the additions in 1-i+2i.

$$2z^{2}+2z=2iz+1+i$$

Subtract 2iz from both sides.

$$2z^{2}+2z-2iz=1+i$$

Combine 2z and -2iz to get \left(2-2i\right)z.

$$2z^{2}+\left(2-2i\right)z=1+i$$

Divide both sides by 2.

$$\frac{2z^{2}+\left(2-2i\right)z}{2}=\frac{1+i}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide 2-2i by 2.

$$z^{2}+\left(1-i\right)z=\frac{1+i}{2}$$

Divide 1+i by 2.

$$z^{2}+\left(1-i\right)z=\frac{1}{2}+\frac{1}{2}i$$

Divide 1-i, the coefficient of the x term, by 2 to get \frac{1}{2}-\frac{1}{2}i. Then add the square of \frac{1}{2}-\frac{1}{2}i to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}+\left(1-i\right)z+\left(\frac{1}{2}-\frac{1}{2}i\right)^{2}=\frac{1}{2}+\frac{1}{2}i+\left(\frac{1}{2}-\frac{1}{2}i\right)^{2}$$

Square \frac{1}{2}-\frac{1}{2}i.

$$z^{2}+\left(1-i\right)z-\frac{1}{2}i=\frac{1}{2}+\frac{1}{2}i-\frac{1}{2}i$$

Add \frac{1}{2}+\frac{1}{2}i to -\frac{1}{2}i.

$$z^{2}+\left(1-i\right)z-\frac{1}{2}i=\frac{1}{2}$$

Factor z^{2}+\left(1-i\right)z-\frac{1}{2}i. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z+\left(\frac{1}{2}-\frac{1}{2}i\right)\right)^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z+\left(\frac{1}{2}-\frac{1}{2}i\right)\right)^{2}}=\sqrt{\frac{1}{2}}$$

Simplify.

$$z+\left(\frac{1}{2}-\frac{1}{2}i\right)=\frac{\sqrt{2}}{2} z+\left(\frac{1}{2}-\frac{1}{2}i\right)=-\frac{\sqrt{2}}{2}$$

Subtract \frac{1}{2}-\frac{1}{2}i from both sides of the equation.

$$z=\frac{\sqrt{2}}{2}+\left(-\frac{1}{2}+\frac{1}{2}i\right) z=-\frac{1}{2}+\frac{1}{2}i-\frac{\sqrt{2}}{2}$$