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

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

Solve for z

$$z=-\frac{3}{5}+\frac{1}{5}i=-0.6+0.2i$$

Steps for Solving Linear Equation

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

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

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

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

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

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

Multiply i times 2+i.

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

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

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

Reorder the terms.

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

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

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

Multiply both sides by 1+i.

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

Multiply complex numbers -\frac{1}{5}+\frac{2}{5}i and 1+i like you multiply binomials.

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

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

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

Do the multiplications in -\frac{1}{5}-\frac{1}{5}i+\frac{2}{5}i\times 1+\frac{2}{5}\left(-1\right).

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

Combine the real and imaginary parts in -\frac{1}{5}-\frac{1}{5}i+\frac{2}{5}i-\frac{2}{5}.

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

Do the additions in -\frac{1}{5}-\frac{2}{5}+\left(-\frac{1}{5}+\frac{2}{5}\right)i.

$$z=-\frac{3}{5}+\frac{1}{5}i$$