Solve 5+3i/2+5i

$$\frac { 5 + 3 i } { 2 + 5 i }$$

Evaluate

$$\frac{25}{29}-\frac{19}{29}i\approx 0.862068966-0.655172414i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-5i.

$$\frac{\left(5+3i\right)\left(2-5i\right)}{\left(2+5i\right)\left(2-5i\right)}$$

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

$$\frac{\left(5+3i\right)\left(2-5i\right)}{2^{2}-5^{2}i^{2}}$$

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

$$\frac{\left(5+3i\right)\left(2-5i\right)}{29}$$

Multiply complex numbers 5+3i and 2-5i like you multiply binomials.

$$\frac{5\times 2+5\times \left(-5i\right)+3i\times 2+3\left(-5\right)i^{2}}{29}$$

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

$$\frac{5\times 2+5\times \left(-5i\right)+3i\times 2+3\left(-5\right)\left(-1\right)}{29}$$

Do the multiplications in 5\times 2+5\times \left(-5i\right)+3i\times 2+3\left(-5\right)\left(-1\right).

$$\frac{10-25i+6i+15}{29}$$

Combine the real and imaginary parts in 10-25i+6i+15.

$$\frac{10+15+\left(-25+6\right)i}{29}$$

Do the additions in 10+15+\left(-25+6\right)i.

$$\frac{25-19i}{29}$$

Divide 25-19i by 29 to get \frac{25}{29}-\frac{19}{29}i.

$$\frac{25}{29}-\frac{19}{29}i$$

Real Part

$$\frac{25}{29} = 0.8620689655172413$$

Solution Steps

Multiply both numerator and denominator of \frac{5+3i}{2+5i} by the complex conjugate of the denominator, 2-5i.

$$Re(\frac{\left(5+3i\right)\left(2-5i\right)}{\left(2+5i\right)\left(2-5i\right)})$$

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

$$Re(\frac{\left(5+3i\right)\left(2-5i\right)}{2^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{\left(5+3i\right)\left(2-5i\right)}{29})$$

Multiply complex numbers 5+3i and 2-5i like you multiply binomials.

$$Re(\frac{5\times 2+5\times \left(-5i\right)+3i\times 2+3\left(-5\right)i^{2}}{29})$$

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

$$Re(\frac{5\times 2+5\times \left(-5i\right)+3i\times 2+3\left(-5\right)\left(-1\right)}{29})$$

Do the multiplications in 5\times 2+5\times \left(-5i\right)+3i\times 2+3\left(-5\right)\left(-1\right).

$$Re(\frac{10-25i+6i+15}{29})$$

Combine the real and imaginary parts in 10-25i+6i+15.

$$Re(\frac{10+15+\left(-25+6\right)i}{29})$$

Do the additions in 10+15+\left(-25+6\right)i.

$$Re(\frac{25-19i}{29})$$

Divide 25-19i by 29 to get \frac{25}{29}-\frac{19}{29}i.

$$Re(\frac{25}{29}-\frac{19}{29}i)$$

The real part of \frac{25}{29}-\frac{19}{29}i is \frac{25}{29}.

$$\frac{25}{29}$$