Solve z=3+4i/1-i

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

Solve for z

$$z=-\frac{1}{2}+\frac{7}{2}i=-0.5+3.5i$$

Steps for Solving Linear Equation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$z=\frac{-1+7i}{2}$$

Divide -1+7i by 2 to get -\frac{1}{2}+\frac{7}{2}i.

$$z=-\frac{1}{2}+\frac{7}{2}i$$

Assign z

$$z≔-\frac{1}{2}+\frac{7}{2}i$$