Solve (3-4i)quadz=1+2i

$$( 3 - 4 i ) \quad z = 1 + 2 i$$

Solve for z

$$z=-\frac{1}{5}+\frac{2}{5}i=-0.2+0.4i$$

Steps for Solving Linear Equation

Divide both sides by 3-4i.

$$z=\frac{1+2i}{3-4i}$$

Multiply both numerator and denominator of \frac{1+2i}{3-4i} by the complex conjugate of the denominator, 3+4i.

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

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

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

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

$$z=\frac{\left(1+2i\right)\left(3+4i\right)}{25}$$

Multiply complex numbers 1+2i and 3+4i like you multiply binomials.

$$z=\frac{1\times 3+1\times \left(4i\right)+2i\times 3+2\times 4i^{2}}{25}$$

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

$$z=\frac{1\times 3+1\times \left(4i\right)+2i\times 3+2\times 4\left(-1\right)}{25}$$

Do the multiplications in 1\times 3+1\times \left(4i\right)+2i\times 3+2\times 4\left(-1\right).

$$z=\frac{3+4i+6i-8}{25}$$

Combine the real and imaginary parts in 3+4i+6i-8.

$$z=\frac{3-8+\left(4+6\right)i}{25}$$

Do the additions in 3-8+\left(4+6\right)i.

$$z=\frac{-5+10i}{25}$$

Divide -5+10i by 25 to get -\frac{1}{5}+\frac{2}{5}i.

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