Solve (1+i)z=2-i

$$( 1 + i ) z = 2 - i$$

Solve for z

$$z=\frac{1}{2}-\frac{3}{2}i=0.5-1.5i$$

Steps for Solving Linear Equation

Divide both sides by 1+i.

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

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

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

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

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

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

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

Multiply complex numbers 2-i and 1-i like you multiply binomials.

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

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

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

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

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

Combine the real and imaginary parts in 2-2i-i-1.

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

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

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

Divide 1-3i by 2 to get \frac{1}{2}-\frac{3}{2}i.

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