Solve (1+2i)(3+4i)/(3-4i)(3+4i)

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

Evaluate

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

Solution Steps

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.

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

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

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

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

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.

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

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

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

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

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