Solve 2iz-4+2i/2-3+i=1+i

$$\frac { 2 i z - 4 + 2 i } { 2 - 3 + i } = 1 + i$$

Solve for z

$$z=-1-i$$

Steps for Solving Linear Equation

Combine the real and imaginary parts in 2-3+i.

$$\frac{2iz-4+2i}{2-3+i}=1+i$$

Add 2 to -3.

$$\frac{2iz-4+2i}{-1+i}=1+i$$

Divide each term of 2iz-4+2i by -1+i to get \frac{2iz}{-1+i}+\frac{-4}{-1+i}+\frac{2i}{-1+i}.

$$\frac{2iz}{-1+i}+\frac{-4}{-1+i}+\frac{2i}{-1+i}=1+i$$

Divide 2iz by -1+i to get \left(1-i\right)z.

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

Multiply both numerator and denominator of \frac{-4}{-1+i} by the complex conjugate of the denominator, -1-i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply -4 times -1-i.

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

Do the multiplications in -4\left(-1\right)-4\left(-i\right).

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

Divide 4+4i by 2 to get 2+2i.

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

Multiply both numerator and denominator of \frac{2i}{-1+i} by the complex conjugate of the denominator, -1-i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply 2i times -1-i.

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

By definition, i^{2} is -1.

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

Do the multiplications in 2i\left(-1\right)+2\left(-1\right)\left(-1\right). Reorder the terms.

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

Divide 2-2i by 2 to get 1-i.

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

Combine the real and imaginary parts in 2+2i+\left(1-i\right).

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

Do the additions in 2+1+\left(2-1\right)i.

$$\left(1-i\right)z+3+i=1+i$$

Subtract 3 from both sides.

$$\left(1-i\right)z+i=1+i-3$$

Subtract 3 from 1+i by subtracting corresponding real and imaginary parts.

$$\left(1-i\right)z+i=1-3+i$$

Subtract 3 from 1 to get -2.

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

Subtract i from both sides.

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

Combine the real and imaginary parts in numbers -2+i and -i.

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

Add 1 to -1.

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

Divide both sides by 1-i.

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

Multiply both numerator and denominator of \frac{-2}{1-i} by the complex conjugate of the denominator, 1+i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply -2 times 1+i.

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

Divide -2-2i by 2 to get -1-i.

$$z=-1-i$$