Solve z_{1}=(1+i)z+2-3i

$$z _ { 1 } = ( 1 + i ) z + 2 - 3 i$$

Solve for z

$$z=\left(\frac{1}{2}-\frac{1}{2}i\right)z_{1}+\left(\frac{1}{2}+\frac{5}{2}i\right)$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\left(1+i\right)z+2-3i=z_{1}$$

Subtract 2 from both sides.

$$\left(1+i\right)z-3i=z_{1}-2$$

Add 3i to both sides.

$$\left(1+i\right)z=z_{1}-2+3i$$

The equation is in standard form.

$$\left(1+i\right)z=z_{1}+\left(-2+3i\right)$$

Divide both sides by 1+i.

$$\frac{\left(1+i\right)z}{1+i}=\frac{z_{1}+\left(-2+3i\right)}{1+i}$$

Dividing by 1+i undoes the multiplication by 1+i.

$$z=\frac{z_{1}+\left(-2+3i\right)}{1+i}$$

Divide z_{1}+\left(-2+3i\right) by 1+i.

$$z=\left(\frac{1}{2}-\frac{1}{2}i\right)z_{1}+\left(\frac{1}{2}+\frac{5}{2}i\right)$$

Solve for z_1

$$z_{1}=\left(1+i\right)z+\left(2-3i\right)$$