Solve 2+3i/-1+i

$$\frac { 2 + 3 i } { - 1 + i }$$

Evaluate

$$\frac{1}{2}-\frac{5}{2}i=0.5-2.5i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -1-i.

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

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

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

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

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

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

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

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

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

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

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

Combine the real and imaginary parts in -2-2i-3i+3.

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

Do the additions in -2+3+\left(-2-3\right)i.

$$\frac{1-5i}{2}$$

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

$$\frac{1}{2}-\frac{5}{2}i$$

Real Part

$$\frac{1}{2} = 0.5$$

Solution Steps

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

$$Re(\frac{\left(2+3i\right)\left(-1-i\right)}{\left(-1+i\right)\left(-1-i\right)})$$

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

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

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

$$Re(\frac{\left(2+3i\right)\left(-1-i\right)}{2})$$

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

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

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

$$Re(\frac{2\left(-1\right)+2\left(-i\right)+3i\left(-1\right)+3\left(-1\right)\left(-1\right)}{2})$$

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

$$Re(\frac{-2-2i-3i+3}{2})$$

Combine the real and imaginary parts in -2-2i-3i+3.

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

Do the additions in -2+3+\left(-2-3\right)i.

$$Re(\frac{1-5i}{2})$$

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

$$Re(\frac{1}{2}-\frac{5}{2}i)$$

The real part of \frac{1}{2}-\frac{5}{2}i is \frac{1}{2}.

$$\frac{1}{2}$$