Solve =3-2i/1+sqrt{-4}

$$= \frac { 3 - 2 i } { 1 + \sqrt { - 4 } }$$

Evaluate

$$-\frac{1}{5}-\frac{8}{5}i=-0.2-1.6i$$

Solution Steps

Calculate the square root of -4 and get 2i.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-2i.

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

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

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

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

$$\frac{\left(3-2i\right)\left(1-2i\right)}{5}$$

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

$$\frac{3\times 1+3\times \left(-2i\right)-2i-2\left(-2\right)i^{2}}{5}$$

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

$$\frac{3\times 1+3\times \left(-2i\right)-2i-2\left(-2\right)\left(-1\right)}{5}$$

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

$$\frac{3-6i-2i-4}{5}$$

Combine the real and imaginary parts in 3-6i-2i-4.

$$\frac{3-4+\left(-6-2\right)i}{5}$$

Do the additions in 3-4+\left(-6-2\right)i.

$$\frac{-1-8i}{5}$$

Divide -1-8i by 5 to get -\frac{1}{5}-\frac{8}{5}i.

$$-\frac{1}{5}-\frac{8}{5}i$$

Real Part

$$-\frac{1}{5} = -0.2$$

Solution Steps

Calculate the square root of -4 and get 2i.

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

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

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

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

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

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

$$Re(\frac{\left(3-2i\right)\left(1-2i\right)}{5})$$

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

$$Re(\frac{3\times 1+3\times \left(-2i\right)-2i-2\left(-2\right)i^{2}}{5})$$

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

$$Re(\frac{3\times 1+3\times \left(-2i\right)-2i-2\left(-2\right)\left(-1\right)}{5})$$

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

$$Re(\frac{3-6i-2i-4}{5})$$

Combine the real and imaginary parts in 3-6i-2i-4.

$$Re(\frac{3-4+\left(-6-2\right)i}{5})$$

Do the additions in 3-4+\left(-6-2\right)i.

$$Re(\frac{-1-8i}{5})$$

Divide -1-8i by 5 to get -\frac{1}{5}-\frac{8}{5}i.

$$Re(-\frac{1}{5}-\frac{8}{5}i)$$

The real part of -\frac{1}{5}-\frac{8}{5}i is -\frac{1}{5}.

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