Solve z/2+ai=2/1+i

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

Solve for a

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

$$z\neq 0$$

Steps for Solving Linear Equation

Variable a cannot be equal to 2i since division by zero is not defined. Multiply both sides of the equation by ia+2.

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

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

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

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

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

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

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

Use the distributive property to multiply ia+2 by 1-i.

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

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

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

Subtract 2-2i from both sides.

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

Multiply -1 and 2-2i to get -2+2i.

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

Divide both sides by 1+i.

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

Dividing by 1+i undoes the multiplication by 1+i.

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

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

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

Variable a cannot be equal to 2i.

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

Solve for z

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

$$a\neq 2i$$

Solution Steps

Multiply both sides of the equation by ia+2.

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

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

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

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

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

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

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

Use the distributive property to multiply ia+2 by 1-i.

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