Solve (a+2i)(1-i)/(1+i)(1-i)=z

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

Solve for a

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

Steps for Solving Linear Equation

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{a+2i}{\left(1+i\right)\left(1-i\right)^{0}}=z$$

Calculate 1-i to the power of 0 and get 1.

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

Multiply 1+i and 1 to get 1+i.

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

Divide each term of a+2i by 1+i to get \frac{a}{1+i}+\frac{2i}{1+i}.

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

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

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

Do the multiplications in \frac{2i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

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

Divide 2+2i by 2 to get 1+i.

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

Subtract 1+i from both sides.

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

Multiply -1 and 1+i to get -1-i.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for z

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