Solve (2-i)z=1+i.ineqj

$$( 2 - i ) z = 1 + i . i \neq j$$

Solve for z

$$z=\frac{1}{5}+\frac{3}{5}i=0.2+0.6i$$

Steps for Solving Linear Equation

Divide both sides by 2-i.

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

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

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

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

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

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

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

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

$$z=\frac{1\times 2+i+2i+i^{2}}{5}$$

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

$$z=\frac{1\times 2+i+2i-1}{5}$$

Do the multiplications in 1\times 2+i+2i-1.

$$z=\frac{2+i+2i-1}{5}$$

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

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

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

$$z=\frac{1+3i}{5}$$

Divide 1+3i by 5 to get \frac{1}{5}+\frac{3}{5}i.

$$z=\frac{1}{5}+\frac{3}{5}i$$