Solve 5-10i/(5+10i)(5-10i)

$$\frac { 5 - 10 i } { ( 5 + 10 i ) ( 5 - 10 i ) }$$

Evaluate

$$\frac{1}{25}-\frac{2}{25}i=0.04-0.08i$$

Solution Steps

Multiply complex numbers 5+10i and 5-10i like you multiply binomials.

$$\frac{5-10i}{5\times 5+5\times \left(-10i\right)+10i\times 5+10\left(-10\right)i^{2}}$$

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

$$\frac{5-10i}{5\times 5+5\times \left(-10i\right)+10i\times 5+10\left(-10\right)\left(-1\right)}$$

Do the multiplications in 5\times 5+5\times \left(-10i\right)+10i\times 5+10\left(-10\right)\left(-1\right).

$$\frac{5-10i}{25-50i+50i+100}$$

Combine the real and imaginary parts in 25-50i+50i+100.

$$\frac{5-10i}{25+100+\left(-50+50\right)i}$$

Do the additions in 25+100+\left(-50+50\right)i.

$$\frac{5-10i}{125}$$

Divide 5-10i by 125 to get \frac{1}{25}-\frac{2}{25}i.

$$\frac{1}{25}-\frac{2}{25}i$$

Real Part

$$\frac{1}{25} = 0.04$$

Solution Steps

Multiply complex numbers 5+10i and 5-10i like you multiply binomials.

$$Re(\frac{5-10i}{5\times 5+5\times \left(-10i\right)+10i\times 5+10\left(-10\right)i^{2}})$$

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

$$Re(\frac{5-10i}{5\times 5+5\times \left(-10i\right)+10i\times 5+10\left(-10\right)\left(-1\right)})$$

Do the multiplications in 5\times 5+5\times \left(-10i\right)+10i\times 5+10\left(-10\right)\left(-1\right).

$$Re(\frac{5-10i}{25-50i+50i+100})$$

Combine the real and imaginary parts in 25-50i+50i+100.

$$Re(\frac{5-10i}{25+100+\left(-50+50\right)i})$$

Do the additions in 25+100+\left(-50+50\right)i.

$$Re(\frac{5-10i}{125})$$

Divide 5-10i by 125 to get \frac{1}{25}-\frac{2}{25}i.

$$Re(\frac{1}{25}-\frac{2}{25}i)$$

The real part of \frac{1}{25}-\frac{2}{25}i is \frac{1}{25}.

$$\frac{1}{25}$$