Solve 240/1+2i

$$\frac{ 240 }{ 1+2i }$$

Evaluate

$$48-96i$$

Solution Steps

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

$$\frac{240\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}$$

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

$$\frac{240\left(1-2i\right)}{1^{2}-2^{2}i^{2}}$$

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

$$\frac{240\left(1-2i\right)}{5}$$

Multiply 240 times 1-2i.

$$\frac{240\times 1+240\times \left(-2i\right)}{5}$$

Do the multiplications in 240\times 1+240\times \left(-2i\right).

$$\frac{240-480i}{5}$$

Divide 240-480i by 5 to get 48-96i.

$$48-96i$$

Real Part

$$48$$

Solution Steps

Multiply both numerator and denominator of \frac{240}{1+2i} by the complex conjugate of the denominator, 1-2i.

$$Re(\frac{240\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)})$$

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

$$Re(\frac{240\left(1-2i\right)}{1^{2}-2^{2}i^{2}})$$

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

$$Re(\frac{240\left(1-2i\right)}{5})$$

Multiply 240 times 1-2i.

$$Re(\frac{240\times 1+240\times \left(-2i\right)}{5})$$

Do the multiplications in 240\times 1+240\times \left(-2i\right).

$$Re(\frac{240-480i}{5})$$

Divide 240-480i by 5 to get 48-96i.

$$Re(48-96i)$$

The real part of 48-96i is 48.

$$48$$