Solve 3+4i/1+4-i/3+2i

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

Evaluate

$$\frac{49}{13}+\frac{41}{13}i\approx 3.769230769+3.153846154i$$

Solution Steps

Anything divided by one gives itself.

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

Multiply both numerator and denominator of \frac{4-i}{3+2i} by the complex conjugate of the denominator, 3-2i.

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

Do the multiplications in \frac{\left(4-i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}.

$$3+4i+\frac{10-11i}{13}$$

Divide 10-11i by 13 to get \frac{10}{13}-\frac{11}{13}i.

$$3+4i+\left(\frac{10}{13}-\frac{11}{13}i\right)$$

Add 3+4i and \frac{10}{13}-\frac{11}{13}i to get \frac{49}{13}+\frac{41}{13}i.

$$\frac{49}{13}+\frac{41}{13}i$$

Real Part

$$\frac{49}{13} = 3\frac{10}{13} = 3.769230769230769$$

Solution Steps

Anything divided by one gives itself.

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

Multiply both numerator and denominator of \frac{4-i}{3+2i} by the complex conjugate of the denominator, 3-2i.

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

Do the multiplications in \frac{\left(4-i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}.

$$Re(3+4i+\frac{10-11i}{13})$$

Divide 10-11i by 13 to get \frac{10}{13}-\frac{11}{13}i.

$$Re(3+4i+\left(\frac{10}{13}-\frac{11}{13}i\right))$$

Add 3+4i and \frac{10}{13}-\frac{11}{13}i to get \frac{49}{13}+\frac{41}{13}i.

$$Re(\frac{49}{13}+\frac{41}{13}i)$$

The real part of \frac{49}{13}+\frac{41}{13}i is \frac{49}{13}.

$$\frac{49}{13}$$