Solve /2+ai=2/1+i

$$\frac{ }{ 2+ai } = \frac{ 2 }{ 1+i }$$

Solve for a

$$a=\frac{1}{2}+\frac{3}{2}i=0.5+1.5i$$

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.

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

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

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

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

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

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

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

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

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

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

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

Subtract 2-2i from both sides.

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

Subtract 2-2i from 1 to get -1+2i.

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

Divide both sides by 1+i.

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

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

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

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

$$a=\frac{1+3i}{2}$$

Divide 1+3i by 2 to get \frac{1}{2}+\frac{3}{2}i.

$$a=\frac{1}{2}+\frac{3}{2}i$$