Solve 5-i/1-ii+9+5i/3+i

$$\frac { 5 - i } { 1 - i i } + \frac { 9 + 5 i } { 3 + i }$$

Evaluate

$$\frac{57}{10}+\frac{1}{10}i=5.7+0.1i$$

Solution Steps

Multiply i and i to get -1.

$$\frac{5-i}{1-\left(-1\right)}+\frac{9+5i}{3+i}$$

The opposite of -1 is 1.

$$\frac{5-i}{1+1}+\frac{9+5i}{3+i}$$

Add 1 and 1 to get 2.

$$\frac{5-i}{2}+\frac{9+5i}{3+i}$$

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

$$\frac{5}{2}-\frac{1}{2}i+\frac{9+5i}{3+i}$$

Multiply both numerator and denominator of \frac{9+5i}{3+i} by the complex conjugate of the denominator, 3-i.

$$\frac{5}{2}-\frac{1}{2}i+\frac{\left(9+5i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$$

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

$$\frac{5}{2}-\frac{1}{2}i+\frac{\left(9+5i\right)\left(3-i\right)}{3^{2}-i^{2}}$$

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

$$\frac{5}{2}-\frac{1}{2}i+\frac{\left(9+5i\right)\left(3-i\right)}{10}$$

Multiply complex numbers 9+5i and 3-i like you multiply binomials.

$$\frac{5}{2}-\frac{1}{2}i+\frac{9\times 3+9\left(-i\right)+5i\times 3+5\left(-1\right)i^{2}}{10}$$

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

$$\frac{5}{2}-\frac{1}{2}i+\frac{9\times 3+9\left(-i\right)+5i\times 3+5\left(-1\right)\left(-1\right)}{10}$$

Do the multiplications in 9\times 3+9\left(-i\right)+5i\times 3+5\left(-1\right)\left(-1\right).

$$\frac{5}{2}-\frac{1}{2}i+\frac{27-9i+15i+5}{10}$$

Combine the real and imaginary parts in 27-9i+15i+5.

$$\frac{5}{2}-\frac{1}{2}i+\frac{27+5+\left(-9+15\right)i}{10}$$

Do the additions in 27+5+\left(-9+15\right)i.

$$\frac{5}{2}-\frac{1}{2}i+\frac{32+6i}{10}$$

Divide 32+6i by 10 to get \frac{16}{5}+\frac{3}{5}i.

$$\frac{5}{2}-\frac{1}{2}i+\left(\frac{16}{5}+\frac{3}{5}i\right)$$

Combine the real and imaginary parts in numbers \frac{5}{2}-\frac{1}{2}i and \frac{16}{5}+\frac{3}{5}i.

$$\frac{5}{2}+\frac{16}{5}+\left(-\frac{1}{2}+\frac{3}{5}\right)i$$

Add \frac{5}{2} to \frac{16}{5}. Add -\frac{1}{2} to \frac{3}{5}.

$$\frac{57}{10}+\frac{1}{10}i$$

Real Part

$$\frac{57}{10} = 5\frac{7}{10} = 5.7$$

Solution Steps

Multiply i and i to get -1.

$$Re(\frac{5-i}{1-\left(-1\right)}+\frac{9+5i}{3+i})$$

The opposite of -1 is 1.

$$Re(\frac{5-i}{1+1}+\frac{9+5i}{3+i})$$

Add 1 and 1 to get 2.

$$Re(\frac{5-i}{2}+\frac{9+5i}{3+i})$$

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

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{9+5i}{3+i})$$

Multiply both numerator and denominator of \frac{9+5i}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{\left(9+5i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)})$$

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

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{\left(9+5i\right)\left(3-i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{\left(9+5i\right)\left(3-i\right)}{10})$$

Multiply complex numbers 9+5i and 3-i like you multiply binomials.

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{9\times 3+9\left(-i\right)+5i\times 3+5\left(-1\right)i^{2}}{10})$$

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

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{9\times 3+9\left(-i\right)+5i\times 3+5\left(-1\right)\left(-1\right)}{10})$$

Do the multiplications in 9\times 3+9\left(-i\right)+5i\times 3+5\left(-1\right)\left(-1\right).

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{27-9i+15i+5}{10})$$

Combine the real and imaginary parts in 27-9i+15i+5.

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{27+5+\left(-9+15\right)i}{10})$$

Do the additions in 27+5+\left(-9+15\right)i.

$$Re(\frac{5}{2}-\frac{1}{2}i+\frac{32+6i}{10})$$

Divide 32+6i by 10 to get \frac{16}{5}+\frac{3}{5}i.

$$Re(\frac{5}{2}-\frac{1}{2}i+\left(\frac{16}{5}+\frac{3}{5}i\right))$$

Combine the real and imaginary parts in numbers \frac{5}{2}-\frac{1}{2}i and \frac{16}{5}+\frac{3}{5}i.

$$Re(\frac{5}{2}+\frac{16}{5}+\left(-\frac{1}{2}+\frac{3}{5}\right)i)$$

Add \frac{5}{2} to \frac{16}{5}. Add -\frac{1}{2} to \frac{3}{5}.

$$Re(\frac{57}{10}+\frac{1}{10}i)$$

The real part of \frac{57}{10}+\frac{1}{10}i is \frac{57}{10}.

$$\frac{57}{10}$$