Solve 2+9i/-3-i

$$\frac { 2 + 9 i } { - 3 - i }$$

Evaluate

$$-\frac{3}{2}-\frac{5}{2}i=-1.5-2.5i$$

Solution Steps

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

$$\frac{\left(2+9i\right)\left(-3+i\right)}{\left(-3-i\right)\left(-3+i\right)}$$

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

$$\frac{\left(2+9i\right)\left(-3+i\right)}{\left(-3\right)^{2}-i^{2}}$$

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

$$\frac{\left(2+9i\right)\left(-3+i\right)}{10}$$

Multiply complex numbers 2+9i and -3+i like you multiply binomials.

$$\frac{2\left(-3\right)+2i+9i\left(-3\right)+9i^{2}}{10}$$

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

$$\frac{2\left(-3\right)+2i+9i\left(-3\right)+9\left(-1\right)}{10}$$

Do the multiplications in 2\left(-3\right)+2i+9i\left(-3\right)+9\left(-1\right).

$$\frac{-6+2i-27i-9}{10}$$

Combine the real and imaginary parts in -6+2i-27i-9.

$$\frac{-6-9+\left(2-27\right)i}{10}$$

Do the additions in -6-9+\left(2-27\right)i.

$$\frac{-15-25i}{10}$$

Divide -15-25i by 10 to get -\frac{3}{2}-\frac{5}{2}i.

$$-\frac{3}{2}-\frac{5}{2}i$$

Real Part

$$-\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

Multiply both numerator and denominator of \frac{2+9i}{-3-i} by the complex conjugate of the denominator, -3+i.

$$Re(\frac{\left(2+9i\right)\left(-3+i\right)}{\left(-3-i\right)\left(-3+i\right)})$$

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

$$Re(\frac{\left(2+9i\right)\left(-3+i\right)}{\left(-3\right)^{2}-i^{2}})$$

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

$$Re(\frac{\left(2+9i\right)\left(-3+i\right)}{10})$$

Multiply complex numbers 2+9i and -3+i like you multiply binomials.

$$Re(\frac{2\left(-3\right)+2i+9i\left(-3\right)+9i^{2}}{10})$$

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

$$Re(\frac{2\left(-3\right)+2i+9i\left(-3\right)+9\left(-1\right)}{10})$$

Do the multiplications in 2\left(-3\right)+2i+9i\left(-3\right)+9\left(-1\right).

$$Re(\frac{-6+2i-27i-9}{10})$$

Combine the real and imaginary parts in -6+2i-27i-9.

$$Re(\frac{-6-9+\left(2-27\right)i}{10})$$

Do the additions in -6-9+\left(2-27\right)i.

$$Re(\frac{-15-25i}{10})$$

Divide -15-25i by 10 to get -\frac{3}{2}-\frac{5}{2}i.

$$Re(-\frac{3}{2}-\frac{5}{2}i)$$

The real part of -\frac{3}{2}-\frac{5}{2}i is -\frac{3}{2}.

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