Solve -4+3i/-10+7i

$$\frac { - 4 + 3 i } { - 10 + 7 i }$$

Evaluate

$$\frac{61}{149}-\frac{2}{149}i\approx 0.409395973-0.013422819i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -10-7i.

$$\frac{\left(-4+3i\right)\left(-10-7i\right)}{\left(-10+7i\right)\left(-10-7i\right)}$$

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

$$\frac{\left(-4+3i\right)\left(-10-7i\right)}{\left(-10\right)^{2}-7^{2}i^{2}}$$

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

$$\frac{\left(-4+3i\right)\left(-10-7i\right)}{149}$$

Multiply complex numbers -4+3i and -10-7i like you multiply binomials.

$$\frac{-4\left(-10\right)-4\times \left(-7i\right)+3i\left(-10\right)+3\left(-7\right)i^{2}}{149}$$

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

$$\frac{-4\left(-10\right)-4\times \left(-7i\right)+3i\left(-10\right)+3\left(-7\right)\left(-1\right)}{149}$$

Do the multiplications in -4\left(-10\right)-4\times \left(-7i\right)+3i\left(-10\right)+3\left(-7\right)\left(-1\right).

$$\frac{40+28i-30i+21}{149}$$

Combine the real and imaginary parts in 40+28i-30i+21.

$$\frac{40+21+\left(28-30\right)i}{149}$$

Do the additions in 40+21+\left(28-30\right)i.

$$\frac{61-2i}{149}$$

Divide 61-2i by 149 to get \frac{61}{149}-\frac{2}{149}i.

$$\frac{61}{149}-\frac{2}{149}i$$

Real Part

$$\frac{61}{149} = 0.40939597315436244$$

Solution Steps

Multiply both numerator and denominator of \frac{-4+3i}{-10+7i} by the complex conjugate of the denominator, -10-7i.

$$Re(\frac{\left(-4+3i\right)\left(-10-7i\right)}{\left(-10+7i\right)\left(-10-7i\right)})$$

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

$$Re(\frac{\left(-4+3i\right)\left(-10-7i\right)}{\left(-10\right)^{2}-7^{2}i^{2}})$$

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

$$Re(\frac{\left(-4+3i\right)\left(-10-7i\right)}{149})$$

Multiply complex numbers -4+3i and -10-7i like you multiply binomials.

$$Re(\frac{-4\left(-10\right)-4\times \left(-7i\right)+3i\left(-10\right)+3\left(-7\right)i^{2}}{149})$$

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

$$Re(\frac{-4\left(-10\right)-4\times \left(-7i\right)+3i\left(-10\right)+3\left(-7\right)\left(-1\right)}{149})$$

Do the multiplications in -4\left(-10\right)-4\times \left(-7i\right)+3i\left(-10\right)+3\left(-7\right)\left(-1\right).

$$Re(\frac{40+28i-30i+21}{149})$$

Combine the real and imaginary parts in 40+28i-30i+21.

$$Re(\frac{40+21+\left(28-30\right)i}{149})$$

Do the additions in 40+21+\left(28-30\right)i.

$$Re(\frac{61-2i}{149})$$

Divide 61-2i by 149 to get \frac{61}{149}-\frac{2}{149}i.

$$Re(\frac{61}{149}-\frac{2}{149}i)$$

The real part of \frac{61}{149}-\frac{2}{149}i is \frac{61}{149}.

$$\frac{61}{149}$$