Solve 1-2i/3-4i

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

Evaluate

$$\frac{11}{25}-\frac{2}{25}i=0.44-0.08i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+4i.

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

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

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

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

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

Multiply complex numbers 1-2i and 3+4i like you multiply binomials.

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

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

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

Do the multiplications in 1\times 3+1\times \left(4i\right)-2i\times 3-2\times 4\left(-1\right).

$$\frac{3+4i-6i+8}{25}$$

Combine the real and imaginary parts in 3+4i-6i+8.

$$\frac{3+8+\left(4-6\right)i}{25}$$

Do the additions in 3+8+\left(4-6\right)i.

$$\frac{11-2i}{25}$$

Divide 11-2i by 25 to get \frac{11}{25}-\frac{2}{25}i.

$$\frac{11}{25}-\frac{2}{25}i$$

Real Part

$$\frac{11}{25} = 0.44$$

Solution Steps

Multiply both numerator and denominator of \frac{1-2i}{3-4i} by the complex conjugate of the denominator, 3+4i.

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

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

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

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

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

Multiply complex numbers 1-2i and 3+4i like you multiply binomials.

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

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

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

Do the multiplications in 1\times 3+1\times \left(4i\right)-2i\times 3-2\times 4\left(-1\right).

$$Re(\frac{3+4i-6i+8}{25})$$

Combine the real and imaginary parts in 3+4i-6i+8.

$$Re(\frac{3+8+\left(4-6\right)i}{25})$$

Do the additions in 3+8+\left(4-6\right)i.

$$Re(\frac{11-2i}{25})$$

Divide 11-2i by 25 to get \frac{11}{25}-\frac{2}{25}i.

$$Re(\frac{11}{25}-\frac{2}{25}i)$$

The real part of \frac{11}{25}-\frac{2}{25}i is \frac{11}{25}.

$$\frac{11}{25}$$