Solve 3-i/left(2-iright)left(5+2iright)

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

Evaluate

$$\frac{37}{145}-\frac{9}{145}i\approx 0.255172414-0.062068966i$$

Solution Steps

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

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

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

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

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

$$\frac{3-i}{10+4i-5i+2}$$

Combine the real and imaginary parts in 10+4i-5i+2.

$$\frac{3-i}{10+2+\left(4-5\right)i}$$

Do the additions in 10+2+\left(4-5\right)i.

$$\frac{3-i}{12-i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 12+i.

$$\frac{\left(3-i\right)\left(12+i\right)}{\left(12-i\right)\left(12+i\right)}$$

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

$$\frac{\left(3-i\right)\left(12+i\right)}{12^{2}-i^{2}}$$

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

$$\frac{\left(3-i\right)\left(12+i\right)}{145}$$

Multiply complex numbers 3-i and 12+i like you multiply binomials.

$$\frac{3\times 12+3i-i\times 12-i^{2}}{145}$$

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

$$\frac{3\times 12+3i-i\times 12-\left(-1\right)}{145}$$

Do the multiplications in 3\times 12+3i-i\times 12-\left(-1\right).

$$\frac{36+3i-12i+1}{145}$$

Combine the real and imaginary parts in 36+3i-12i+1.

$$\frac{36+1+\left(3-12\right)i}{145}$$

Do the additions in 36+1+\left(3-12\right)i.

$$\frac{37-9i}{145}$$

Divide 37-9i by 145 to get \frac{37}{145}-\frac{9}{145}i.

$$\frac{37}{145}-\frac{9}{145}i$$

Real Part

$$\frac{37}{145} = 0.25517241379310346$$

Solution Steps

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

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

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

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

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

$$Re(\frac{3-i}{10+4i-5i+2})$$

Combine the real and imaginary parts in 10+4i-5i+2.

$$Re(\frac{3-i}{10+2+\left(4-5\right)i})$$

Do the additions in 10+2+\left(4-5\right)i.

$$Re(\frac{3-i}{12-i})$$

Multiply both numerator and denominator of \frac{3-i}{12-i} by the complex conjugate of the denominator, 12+i.

$$Re(\frac{\left(3-i\right)\left(12+i\right)}{\left(12-i\right)\left(12+i\right)})$$

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

$$Re(\frac{\left(3-i\right)\left(12+i\right)}{12^{2}-i^{2}})$$

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

$$Re(\frac{\left(3-i\right)\left(12+i\right)}{145})$$

Multiply complex numbers 3-i and 12+i like you multiply binomials.

$$Re(\frac{3\times 12+3i-i\times 12-i^{2}}{145})$$

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

$$Re(\frac{3\times 12+3i-i\times 12-\left(-1\right)}{145})$$

Do the multiplications in 3\times 12+3i-i\times 12-\left(-1\right).

$$Re(\frac{36+3i-12i+1}{145})$$

Combine the real and imaginary parts in 36+3i-12i+1.

$$Re(\frac{36+1+\left(3-12\right)i}{145})$$

Do the additions in 36+1+\left(3-12\right)i.

$$Re(\frac{37-9i}{145})$$

Divide 37-9i by 145 to get \frac{37}{145}-\frac{9}{145}i.

$$Re(\frac{37}{145}-\frac{9}{145}i)$$

The real part of \frac{37}{145}-\frac{9}{145}i is \frac{37}{145}.

$$\frac{37}{145}$$