Solve z=(a-i/2+i)

$$z = ( \frac { a - i } { 2 + i } )$$

Solve for a

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

Steps for Solving Linear Equation

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

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

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

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

Do the multiplications in \frac{-i\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}.

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

Divide -1-2i by 5 to get -\frac{1}{5}-\frac{2}{5}i.

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

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

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

Add \frac{1}{5}+\frac{2}{5}i to both sides.

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

The equation is in standard form.

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

Divide both sides by \frac{2}{5}-\frac{1}{5}i.

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

Dividing by \frac{2}{5}-\frac{1}{5}i undoes the multiplication by \frac{2}{5}-\frac{1}{5}i.

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

Divide z+\left(\frac{1}{5}+\frac{2}{5}i\right) by \frac{2}{5}-\frac{1}{5}i.

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

Solve for z

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