Solve -6-17i/2-3i

$$\frac { - 6 - 17 i } { 2 - 3 i }$$

Evaluate

$$3-4i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2+3i.

$$\frac{\left(-6-17i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}$$

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

$$\frac{\left(-6-17i\right)\left(2+3i\right)}{2^{2}-3^{2}i^{2}}$$

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

$$\frac{\left(-6-17i\right)\left(2+3i\right)}{13}$$

Multiply complex numbers -6-17i and 2+3i like you multiply binomials.

$$\frac{-6\times 2-6\times \left(3i\right)-17i\times 2-17\times 3i^{2}}{13}$$

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

$$\frac{-6\times 2-6\times \left(3i\right)-17i\times 2-17\times 3\left(-1\right)}{13}$$

Do the multiplications in -6\times 2-6\times \left(3i\right)-17i\times 2-17\times 3\left(-1\right).

$$\frac{-12-18i-34i+51}{13}$$

Combine the real and imaginary parts in -12-18i-34i+51.

$$\frac{-12+51+\left(-18-34\right)i}{13}$$

Do the additions in -12+51+\left(-18-34\right)i.

$$\frac{39-52i}{13}$$

Divide 39-52i by 13 to get 3-4i.

$$3-4i$$

Real Part

$$3$$

Solution Steps

Multiply both numerator and denominator of \frac{-6-17i}{2-3i} by the complex conjugate of the denominator, 2+3i.

$$Re(\frac{\left(-6-17i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)})$$

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

$$Re(\frac{\left(-6-17i\right)\left(2+3i\right)}{2^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{\left(-6-17i\right)\left(2+3i\right)}{13})$$

Multiply complex numbers -6-17i and 2+3i like you multiply binomials.

$$Re(\frac{-6\times 2-6\times \left(3i\right)-17i\times 2-17\times 3i^{2}}{13})$$

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

$$Re(\frac{-6\times 2-6\times \left(3i\right)-17i\times 2-17\times 3\left(-1\right)}{13})$$

Do the multiplications in -6\times 2-6\times \left(3i\right)-17i\times 2-17\times 3\left(-1\right).

$$Re(\frac{-12-18i-34i+51}{13})$$

Combine the real and imaginary parts in -12-18i-34i+51.

$$Re(\frac{-12+51+\left(-18-34\right)i}{13})$$

Do the additions in -12+51+\left(-18-34\right)i.

$$Re(\frac{39-52i}{13})$$

Divide 39-52i by 13 to get 3-4i.

$$Re(3-4i)$$

The real part of 3-4i is 3.

$$3$$