Solve (1+i)z+5=2-3i

$$( 1 + i ) z + 5 = 2 - 3 i$$

Solve for z

$$z=-3$$

Steps for Solving Linear Equation

Subtract 5 from both sides.

$$\left(1+i\right)z=2-3i-5$$

Subtract 5 from 2-3i by subtracting corresponding real and imaginary parts.

$$\left(1+i\right)z=2-5-3i$$

Subtract 5 from 2 to get -3.

$$\left(1+i\right)z=-3-3i$$

Divide both sides by 1+i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$z=\frac{-6}{2}$$

Divide -6 by 2 to get -3.

$$z=-3$$