Solve 1+i/z=1-i

$$\frac { 1 + i } { z } = 1 - i$$

Solve for z

$$z=i$$

Steps for Solving Linear Equation

Variable z cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by z.

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

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

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

Divide both sides by 1-i.

$$z=\frac{1+i}{1-i}$$

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

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

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

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

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

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

Multiply complex numbers 1+i and 1+i like you multiply binomials.

$$z=\frac{1\times 1+i+i+i^{2}}{2}$$

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

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

Do the multiplications in 1\times 1+i+i-1.

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

Combine the real and imaginary parts in 1+i+i-1.

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

Do the additions in 1-1+\left(1+1\right)i.

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

Divide 2i by 2 to get i.

$$z=i$$