Solve -1+6i/-12+42i

$$\frac { - 1 + 6 i } { - 12 + 42 i }$$

Evaluate

$$\frac{22}{159}-\frac{5}{318}i\approx 0.13836478-0.01572327i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -12-42i.

$$\frac{\left(-1+6i\right)\left(-12-42i\right)}{\left(-12+42i\right)\left(-12-42i\right)}$$

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

$$\frac{\left(-1+6i\right)\left(-12-42i\right)}{\left(-12\right)^{2}-42^{2}i^{2}}$$

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

$$\frac{\left(-1+6i\right)\left(-12-42i\right)}{1908}$$

Multiply complex numbers -1+6i and -12-42i like you multiply binomials.

$$\frac{-\left(-12\right)-\left(-42i\right)+6i\left(-12\right)+6\left(-42\right)i^{2}}{1908}$$

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

$$\frac{-\left(-12\right)-\left(-42i\right)+6i\left(-12\right)+6\left(-42\right)\left(-1\right)}{1908}$$

Do the multiplications in -\left(-12\right)-\left(-42i\right)+6i\left(-12\right)+6\left(-42\right)\left(-1\right).

$$\frac{12+42i-72i+252}{1908}$$

Combine the real and imaginary parts in 12+42i-72i+252.

$$\frac{12+252+\left(42-72\right)i}{1908}$$

Do the additions in 12+252+\left(42-72\right)i.

$$\frac{264-30i}{1908}$$

Divide 264-30i by 1908 to get \frac{22}{159}-\frac{5}{318}i.

$$\frac{22}{159}-\frac{5}{318}i$$

Real Part

$$\frac{22}{159} = 0.13836477987421383$$

Solution Steps

Multiply both numerator and denominator of \frac{-1+6i}{-12+42i} by the complex conjugate of the denominator, -12-42i.

$$Re(\frac{\left(-1+6i\right)\left(-12-42i\right)}{\left(-12+42i\right)\left(-12-42i\right)})$$

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

$$Re(\frac{\left(-1+6i\right)\left(-12-42i\right)}{\left(-12\right)^{2}-42^{2}i^{2}})$$

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

$$Re(\frac{\left(-1+6i\right)\left(-12-42i\right)}{1908})$$

Multiply complex numbers -1+6i and -12-42i like you multiply binomials.

$$Re(\frac{-\left(-12\right)-\left(-42i\right)+6i\left(-12\right)+6\left(-42\right)i^{2}}{1908})$$

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

$$Re(\frac{-\left(-12\right)-\left(-42i\right)+6i\left(-12\right)+6\left(-42\right)\left(-1\right)}{1908})$$

Do the multiplications in -\left(-12\right)-\left(-42i\right)+6i\left(-12\right)+6\left(-42\right)\left(-1\right).

$$Re(\frac{12+42i-72i+252}{1908})$$

Combine the real and imaginary parts in 12+42i-72i+252.

$$Re(\frac{12+252+\left(42-72\right)i}{1908})$$

Do the additions in 12+252+\left(42-72\right)i.

$$Re(\frac{264-30i}{1908})$$

Divide 264-30i by 1908 to get \frac{22}{159}-\frac{5}{318}i.

$$Re(\frac{22}{159}-\frac{5}{318}i)$$

The real part of \frac{22}{159}-\frac{5}{318}i is \frac{22}{159}.

$$\frac{22}{159}$$