Solve 2/5z-i=i/5z-1

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

Solve for z

$$z=-3+i$$

Steps for Solving Linear Equation

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

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

Subtract \frac{1}{5}iz from both sides.

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

Combine \frac{2}{5}z and -\frac{1}{5}iz to get \left(\frac{2}{5}-\frac{1}{5}i\right)z.

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

Add i to both sides.

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

Divide both sides by \frac{2}{5}-\frac{1}{5}i.

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

Multiply both numerator and denominator of \frac{-1+i}{\frac{2}{5}-\frac{1}{5}i} by the complex conjugate of the denominator, \frac{2}{5}+\frac{1}{5}i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide -\frac{3}{5}+\frac{1}{5}i by \frac{1}{5} to get -3+i.

$$z=-3+i$$