Solve 8+61/3+4i

$$\frac { 8 + 61 } { 3 + 4 i }$$

Evaluate

$$\frac{207}{25}-\frac{276}{25}i=8.28-11.04i$$

Solution Steps

Add 8 and 61 to get 69.

$$\frac{69}{3+4i}$$

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

$$\frac{69\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}$$

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

$$\frac{69\left(3-4i\right)}{3^{2}-4^{2}i^{2}}$$

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

$$\frac{69\left(3-4i\right)}{25}$$

Multiply 69 times 3-4i.

$$\frac{69\times 3+69\times \left(-4i\right)}{25}$$

Do the multiplications in 69\times 3+69\times \left(-4i\right).

$$\frac{207-276i}{25}$$

Divide 207-276i by 25 to get \frac{207}{25}-\frac{276}{25}i.

$$\frac{207}{25}-\frac{276}{25}i$$

Real Part

$$\frac{207}{25} = 8\frac{7}{25} = 8.28$$

Solution Steps

Add 8 and 61 to get 69.

$$Re(\frac{69}{3+4i})$$

Multiply both numerator and denominator of \frac{69}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$Re(\frac{69\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)})$$

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

$$Re(\frac{69\left(3-4i\right)}{3^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{69\left(3-4i\right)}{25})$$

Multiply 69 times 3-4i.

$$Re(\frac{69\times 3+69\times \left(-4i\right)}{25})$$

Do the multiplications in 69\times 3+69\times \left(-4i\right).

$$Re(\frac{207-276i}{25})$$

Divide 207-276i by 25 to get \frac{207}{25}-\frac{276}{25}i.

$$Re(\frac{207}{25}-\frac{276}{25}i)$$

The real part of \frac{207}{25}-\frac{276}{25}i is \frac{207}{25}.

$$\frac{207}{25}$$