Solve z=-5+10z/3+4i

$$z = \frac { - 5 + 10 z } { 3 + 4 i }$$

Solve for z

$$z=\frac{7}{13}+\frac{4}{13}i\approx 0.538461538+0.307692308i$$

Steps for Solving Linear Equation

Divide each term of -5+10z by 3+4i to get \frac{-5}{3+4i}+\frac{10z}{3+4i}.

$$z=\frac{-5}{3+4i}+\frac{10z}{3+4i}$$

Multiply both numerator and denominator of \frac{-5}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$z=\frac{-5\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}+\frac{10z}{3+4i}$$

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

$$z=\frac{-5\left(3-4i\right)}{3^{2}-4^{2}i^{2}}+\frac{10z}{3+4i}$$

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

$$z=\frac{-5\left(3-4i\right)}{25}+\frac{10z}{3+4i}$$

Multiply -5 times 3-4i.

$$z=\frac{-5\times 3-5\times \left(-4i\right)}{25}+\frac{10z}{3+4i}$$

Do the multiplications in -5\times 3-5\times \left(-4i\right).

$$z=\frac{-15+20i}{25}+\frac{10z}{3+4i}$$

Divide -15+20i by 25 to get -\frac{3}{5}+\frac{4}{5}i.

$$z=-\frac{3}{5}+\frac{4}{5}i+\frac{10z}{3+4i}$$

Divide 10z by 3+4i to get \left(\frac{6}{5}-\frac{8}{5}i\right)z.

$$z=-\frac{3}{5}+\frac{4}{5}i+\left(\frac{6}{5}-\frac{8}{5}i\right)z$$

Subtract \left(\frac{6}{5}-\frac{8}{5}i\right)z from both sides.

$$z-\left(\frac{6}{5}-\frac{8}{5}i\right)z=-\frac{3}{5}+\frac{4}{5}i$$

Combine z and \left(-\frac{6}{5}+\frac{8}{5}i\right)z to get \left(-\frac{1}{5}+\frac{8}{5}i\right)z.

$$\left(-\frac{1}{5}+\frac{8}{5}i\right)z=-\frac{3}{5}+\frac{4}{5}i$$

Divide both sides by -\frac{1}{5}+\frac{8}{5}i.

$$z=\frac{-\frac{3}{5}+\frac{4}{5}i}{-\frac{1}{5}+\frac{8}{5}i}$$

Multiply both numerator and denominator of \frac{-\frac{3}{5}+\frac{4}{5}i}{-\frac{1}{5}+\frac{8}{5}i} by the complex conjugate of the denominator, -\frac{1}{5}-\frac{8}{5}i.

$$z=\frac{\left(-\frac{3}{5}+\frac{4}{5}i\right)\left(-\frac{1}{5}-\frac{8}{5}i\right)}{\left(-\frac{1}{5}+\frac{8}{5}i\right)\left(-\frac{1}{5}-\frac{8}{5}i\right)}$$

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

$$z=\frac{\left(-\frac{3}{5}+\frac{4}{5}i\right)\left(-\frac{1}{5}-\frac{8}{5}i\right)}{\left(-\frac{1}{5}\right)^{2}-\left(\frac{8}{5}\right)^{2}i^{2}}$$

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

$$z=\frac{\left(-\frac{3}{5}+\frac{4}{5}i\right)\left(-\frac{1}{5}-\frac{8}{5}i\right)}{\frac{13}{5}}$$

Multiply complex numbers -\frac{3}{5}+\frac{4}{5}i and -\frac{1}{5}-\frac{8}{5}i like you multiply binomials.

$$z=\frac{-\frac{3}{5}\left(-\frac{1}{5}\right)-\frac{3}{5}\times \left(-\frac{8}{5}i\right)+\frac{4}{5}i\left(-\frac{1}{5}\right)+\frac{4}{5}\left(-\frac{8}{5}\right)i^{2}}{\frac{13}{5}}$$

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

$$z=\frac{-\frac{3}{5}\left(-\frac{1}{5}\right)-\frac{3}{5}\times \left(-\frac{8}{5}i\right)+\frac{4}{5}i\left(-\frac{1}{5}\right)+\frac{4}{5}\left(-\frac{8}{5}\right)\left(-1\right)}{\frac{13}{5}}$$

Do the multiplications in -\frac{3}{5}\left(-\frac{1}{5}\right)-\frac{3}{5}\times \left(-\frac{8}{5}i\right)+\frac{4}{5}i\left(-\frac{1}{5}\right)+\frac{4}{5}\left(-\frac{8}{5}\right)\left(-1\right).

$$z=\frac{\frac{3}{25}+\frac{24}{25}i-\frac{4}{25}i+\frac{32}{25}}{\frac{13}{5}}$$

Combine the real and imaginary parts in \frac{3}{25}+\frac{24}{25}i-\frac{4}{25}i+\frac{32}{25}.

$$z=\frac{\frac{3}{25}+\frac{32}{25}+\left(\frac{24}{25}-\frac{4}{25}\right)i}{\frac{13}{5}}$$

Do the additions in \frac{3}{25}+\frac{32}{25}+\left(\frac{24}{25}-\frac{4}{25}\right)i.

$$z=\frac{\frac{7}{5}+\frac{4}{5}i}{\frac{13}{5}}$$

Divide \frac{7}{5}+\frac{4}{5}i by \frac{13}{5} to get \frac{7}{13}+\frac{4}{13}i.

$$z=\frac{7}{13}+\frac{4}{13}i$$