Solve 1/frac{1{1+2i}+1/3-4i}

$$\frac { 1 } { \frac { 1 } { 1 + 2 i } + \frac { 1 } { 3 - 4 i } }$$

Evaluate

$$2+\frac{3}{2}i=2+1.5i$$

Solution Steps

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

$$\frac{1}{\frac{1\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}+\frac{1}{3-4i}}$$

Do the multiplications in \frac{1\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$\frac{1}{\frac{1-2i}{5}+\frac{1}{3-4i}}$$

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

$$\frac{1}{\frac{1}{5}-\frac{2}{5}i+\frac{1}{3-4i}}$$

Multiply both numerator and denominator of \frac{1}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$\frac{1}{\frac{1}{5}-\frac{2}{5}i+\frac{1\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}}$$

Do the multiplications in \frac{1\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$\frac{1}{\frac{1}{5}-\frac{2}{5}i+\frac{3+4i}{25}}$$

Divide 3+4i by 25 to get \frac{3}{25}+\frac{4}{25}i.

$$\frac{1}{\frac{1}{5}-\frac{2}{5}i+\left(\frac{3}{25}+\frac{4}{25}i\right)}$$

Add \frac{1}{5}-\frac{2}{5}i and \frac{3}{25}+\frac{4}{25}i to get \frac{8}{25}-\frac{6}{25}i.

$$\frac{1}{\frac{8}{25}-\frac{6}{25}i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, \frac{8}{25}+\frac{6}{25}i.

$$\frac{1\left(\frac{8}{25}+\frac{6}{25}i\right)}{\left(\frac{8}{25}-\frac{6}{25}i\right)\left(\frac{8}{25}+\frac{6}{25}i\right)}$$

Do the multiplications in \frac{1\left(\frac{8}{25}+\frac{6}{25}i\right)}{\left(\frac{8}{25}-\frac{6}{25}i\right)\left(\frac{8}{25}+\frac{6}{25}i\right)}.

$$\frac{\frac{8}{25}+\frac{6}{25}i}{\frac{4}{25}}$$

Divide \frac{8}{25}+\frac{6}{25}i by \frac{4}{25} to get 2+\frac{3}{2}i.

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

Real Part

$$2$$

Solution Steps

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

$$Re(\frac{1}{\frac{1\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}+\frac{1}{3-4i}})$$

Do the multiplications in \frac{1\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$Re(\frac{1}{\frac{1-2i}{5}+\frac{1}{3-4i}})$$

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

$$Re(\frac{1}{\frac{1}{5}-\frac{2}{5}i+\frac{1}{3-4i}})$$

Multiply both numerator and denominator of \frac{1}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$Re(\frac{1}{\frac{1}{5}-\frac{2}{5}i+\frac{1\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}})$$

Do the multiplications in \frac{1\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$Re(\frac{1}{\frac{1}{5}-\frac{2}{5}i+\frac{3+4i}{25}})$$

Divide 3+4i by 25 to get \frac{3}{25}+\frac{4}{25}i.

$$Re(\frac{1}{\frac{1}{5}-\frac{2}{5}i+\left(\frac{3}{25}+\frac{4}{25}i\right)})$$

Add \frac{1}{5}-\frac{2}{5}i and \frac{3}{25}+\frac{4}{25}i to get \frac{8}{25}-\frac{6}{25}i.

$$Re(\frac{1}{\frac{8}{25}-\frac{6}{25}i})$$

Multiply both numerator and denominator of \frac{1}{\frac{8}{25}-\frac{6}{25}i} by the complex conjugate of the denominator, \frac{8}{25}+\frac{6}{25}i.

$$Re(\frac{1\left(\frac{8}{25}+\frac{6}{25}i\right)}{\left(\frac{8}{25}-\frac{6}{25}i\right)\left(\frac{8}{25}+\frac{6}{25}i\right)})$$

Do the multiplications in \frac{1\left(\frac{8}{25}+\frac{6}{25}i\right)}{\left(\frac{8}{25}-\frac{6}{25}i\right)\left(\frac{8}{25}+\frac{6}{25}i\right)}.

$$Re(\frac{\frac{8}{25}+\frac{6}{25}i}{\frac{4}{25}})$$

Divide \frac{8}{25}+\frac{6}{25}i by \frac{4}{25} to get 2+\frac{3}{2}i.

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

The real part of 2+\frac{3}{2}i is 2.

$$2$$