Solve (3+i)*z-i*(2-i)=1

$$( 3 + i ) \cdot z - i \cdot ( 2 - i ) = 1$$

Solve for z

$$z=\frac{4}{5}+\frac{2}{5}i=0.8+0.4i$$

Steps for Solving Linear Equation

Multiply i times 2-i.

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

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

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

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

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

Add 1+2i to both sides.

$$\left(3+i\right)z=1+\left(1+2i\right)$$

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

$$\left(3+i\right)z=1+1+2i$$

Add 1 to 1.

$$\left(3+i\right)z=2+2i$$

Divide both sides by 3+i.

$$z=\frac{2+2i}{3+i}$$

Multiply both numerator and denominator of \frac{2+2i}{3+i} by the complex conjugate of the denominator, 3-i.

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

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

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

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

$$z=\frac{\left(2+2i\right)\left(3-i\right)}{10}$$

Multiply complex numbers 2+2i and 3-i like you multiply binomials.

$$z=\frac{2\times 3+2\left(-i\right)+2i\times 3+2\left(-1\right)i^{2}}{10}$$

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

$$z=\frac{2\times 3+2\left(-i\right)+2i\times 3+2\left(-1\right)\left(-1\right)}{10}$$

Do the multiplications in 2\times 3+2\left(-i\right)+2i\times 3+2\left(-1\right)\left(-1\right).

$$z=\frac{6-2i+6i+2}{10}$$

Combine the real and imaginary parts in 6-2i+6i+2.

$$z=\frac{6+2+\left(-2+6\right)i}{10}$$

Do the additions in 6+2+\left(-2+6\right)i.

$$z=\frac{8+4i}{10}$$

Divide 8+4i by 10 to get \frac{4}{5}+\frac{2}{5}i.

$$z=\frac{4}{5}+\frac{2}{5}i$$