Solve 5-1/3-2i

$$\frac { 5 - 1 } { 3 - 2 i }$$

Evaluate

$$\frac{12}{13}+\frac{8}{13}i\approx 0.923076923+0.615384615i$$

Solution Steps

Subtract 1 from 5 to get 4.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+2i.

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

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

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

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

$$\frac{4\left(3+2i\right)}{13}$$

Multiply 4 times 3+2i.

$$\frac{4\times 3+4\times \left(2i\right)}{13}$$

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

$$\frac{12+8i}{13}$$

Divide 12+8i by 13 to get \frac{12}{13}+\frac{8}{13}i.

$$\frac{12}{13}+\frac{8}{13}i$$

Real Part

$$\frac{12}{13} = 0.9230769230769231$$

Solution Steps

Subtract 1 from 5 to get 4.

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

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

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

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

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

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

$$Re(\frac{4\left(3+2i\right)}{13})$$

Multiply 4 times 3+2i.

$$Re(\frac{4\times 3+4\times \left(2i\right)}{13})$$

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

$$Re(\frac{12+8i}{13})$$

Divide 12+8i by 13 to get \frac{12}{13}+\frac{8}{13}i.

$$Re(\frac{12}{13}+\frac{8}{13}i)$$

The real part of \frac{12}{13}+\frac{8}{13}i is \frac{12}{13}.

$$\frac{12}{13}$$