Solve z=2-bi/1+2i

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

Solve for b

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

Steps for Solving Linear Equation

Divide each term of 2-bi by 1+2i to get \frac{2}{1+2i}-\frac{bi}{1+2i}.

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

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

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

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

$$z=\frac{2-4i}{5}-\frac{bi}{1+2i}$$

Divide 2-4i by 5 to get \frac{2}{5}-\frac{4}{5}i.

$$z=\frac{2}{5}-\frac{4}{5}i-\frac{bi}{1+2i}$$

Divide bi by 1+2i to get b\left(\frac{2}{5}+\frac{1}{5}i\right).

$$z=\frac{2}{5}-\frac{4}{5}i-b\left(\frac{2}{5}+\frac{1}{5}i\right)$$

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

$$\frac{2}{5}-\frac{4}{5}i-b\left(\frac{2}{5}+\frac{1}{5}i\right)=z$$

Subtract \frac{2}{5}-\frac{4}{5}i from both sides.

$$-b\left(\frac{2}{5}+\frac{1}{5}i\right)=z-\left(\frac{2}{5}-\frac{4}{5}i\right)$$

Multiply -1 and \frac{2}{5}+\frac{1}{5}i to get -\frac{2}{5}-\frac{1}{5}i.

$$\left(-\frac{2}{5}-\frac{1}{5}i\right)b=z-\left(\frac{2}{5}-\frac{4}{5}i\right)$$

Multiply -1 and \frac{2}{5}-\frac{4}{5}i to get -\frac{2}{5}+\frac{4}{5}i.

$$\left(-\frac{2}{5}-\frac{1}{5}i\right)b=z+\left(-\frac{2}{5}+\frac{4}{5}i\right)$$

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

$$\frac{\left(-\frac{2}{5}-\frac{1}{5}i\right)b}{-\frac{2}{5}-\frac{1}{5}i}=\frac{z+\left(-\frac{2}{5}+\frac{4}{5}i\right)}{-\frac{2}{5}-\frac{1}{5}i}$$

Dividing by -\frac{2}{5}-\frac{1}{5}i undoes the multiplication by -\frac{2}{5}-\frac{1}{5}i.

$$b=\frac{z+\left(-\frac{2}{5}+\frac{4}{5}i\right)}{-\frac{2}{5}-\frac{1}{5}i}$$

Divide z+\left(-\frac{2}{5}+\frac{4}{5}i\right) by -\frac{2}{5}-\frac{1}{5}i.

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

Solve for z

$$z=\left(-\frac{2}{5}-\frac{1}{5}i\right)b+\left(\frac{2}{5}-\frac{4}{5}i\right)$$

Solution Steps

Divide each term of 2-bi by 1+2i to get \frac{2}{1+2i}-\frac{bi}{1+2i}.

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

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

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

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

$$z=\frac{2-4i}{5}-\frac{bi}{1+2i}$$

Divide 2-4i by 5 to get \frac{2}{5}-\frac{4}{5}i.

$$z=\frac{2}{5}-\frac{4}{5}i-\frac{bi}{1+2i}$$

Divide bi by 1+2i to get b\left(\frac{2}{5}+\frac{1}{5}i\right).

$$z=\frac{2}{5}-\frac{4}{5}i-b\left(\frac{2}{5}+\frac{1}{5}i\right)$$

Multiply -1 and \frac{2}{5}+\frac{1}{5}i to get -\frac{2}{5}-\frac{1}{5}i.

$$z=\frac{2}{5}-\frac{4}{5}i+\left(-\frac{2}{5}-\frac{1}{5}i\right)b$$