Solve -1+2i/5-4i=

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

Evaluate

$$-\frac{13}{41}+\frac{6}{41}i\approx -0.317073171+0.146341463i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5+4i.

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

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

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

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

$$\frac{\left(-1+2i\right)\left(5+4i\right)}{41}$$

Multiply complex numbers -1+2i and 5+4i like you multiply binomials.

$$\frac{-5-4i+2i\times 5+2\times 4i^{2}}{41}$$

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

$$\frac{-5-4i+2i\times 5+2\times 4\left(-1\right)}{41}$$

Do the multiplications in -5-4i+2i\times 5+2\times 4\left(-1\right).

$$\frac{-5-4i+10i-8}{41}$$

Combine the real and imaginary parts in -5-4i+10i-8.

$$\frac{-5-8+\left(-4+10\right)i}{41}$$

Do the additions in -5-8+\left(-4+10\right)i.

$$\frac{-13+6i}{41}$$

Divide -13+6i by 41 to get -\frac{13}{41}+\frac{6}{41}i.

$$-\frac{13}{41}+\frac{6}{41}i$$

Real Part

$$-\frac{13}{41} = -0.3170731707317073$$

Solution Steps

Multiply both numerator and denominator of \frac{-1+2i}{5-4i} by the complex conjugate of the denominator, 5+4i.

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

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

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

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

$$Re(\frac{\left(-1+2i\right)\left(5+4i\right)}{41})$$

Multiply complex numbers -1+2i and 5+4i like you multiply binomials.

$$Re(\frac{-5-4i+2i\times 5+2\times 4i^{2}}{41})$$

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

$$Re(\frac{-5-4i+2i\times 5+2\times 4\left(-1\right)}{41})$$

Do the multiplications in -5-4i+2i\times 5+2\times 4\left(-1\right).

$$Re(\frac{-5-4i+10i-8}{41})$$

Combine the real and imaginary parts in -5-4i+10i-8.

$$Re(\frac{-5-8+\left(-4+10\right)i}{41})$$

Do the additions in -5-8+\left(-4+10\right)i.

$$Re(\frac{-13+6i}{41})$$

Divide -13+6i by 41 to get -\frac{13}{41}+\frac{6}{41}i.

$$Re(-\frac{13}{41}+\frac{6}{41}i)$$

The real part of -\frac{13}{41}+\frac{6}{41}i is -\frac{13}{41}.

$$-\frac{13}{41}$$