Solve 1-2i/3+4i

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

Evaluate

$$-\frac{1}{5}-\frac{2}{5}i=-0.2-0.4i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-4i.

$$\frac{\left(1-2i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}$$

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

$$\frac{\left(1-2i\right)\left(3-4i\right)}{3^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(1-2i\right)\left(3-4i\right)}{25}$$

Multiply complex numbers 1-2i and 3-4i like you multiply binomials.

$$\frac{1\times 3+1\times \left(-4i\right)-2i\times 3-2\left(-4\right)i^{2}}{25}$$

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

$$\frac{1\times 3+1\times \left(-4i\right)-2i\times 3-2\left(-4\right)\left(-1\right)}{25}$$

Do the multiplications in 1\times 3+1\times \left(-4i\right)-2i\times 3-2\left(-4\right)\left(-1\right).

$$\frac{3-4i-6i-8}{25}$$

Combine the real and imaginary parts in 3-4i-6i-8.

$$\frac{3-8+\left(-4-6\right)i}{25}$$

Do the additions in 3-8+\left(-4-6\right)i.

$$\frac{-5-10i}{25}$$

Divide -5-10i by 25 to get -\frac{1}{5}-\frac{2}{5}i.

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

Real Part

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

Solution Steps

Multiply both numerator and denominator of \frac{1-2i}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$Re(\frac{\left(1-2i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)})$$

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

$$Re(\frac{\left(1-2i\right)\left(3-4i\right)}{3^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(1-2i\right)\left(3-4i\right)}{25})$$

Multiply complex numbers 1-2i and 3-4i like you multiply binomials.

$$Re(\frac{1\times 3+1\times \left(-4i\right)-2i\times 3-2\left(-4\right)i^{2}}{25})$$

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

$$Re(\frac{1\times 3+1\times \left(-4i\right)-2i\times 3-2\left(-4\right)\left(-1\right)}{25})$$

Do the multiplications in 1\times 3+1\times \left(-4i\right)-2i\times 3-2\left(-4\right)\left(-1\right).

$$Re(\frac{3-4i-6i-8}{25})$$

Combine the real and imaginary parts in 3-4i-6i-8.

$$Re(\frac{3-8+\left(-4-6\right)i}{25})$$

Do the additions in 3-8+\left(-4-6\right)i.

$$Re(\frac{-5-10i}{25})$$

Divide -5-10i by 25 to get -\frac{1}{5}-\frac{2}{5}i.

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

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

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