Solve 18-3-1/(2-i)(5+20i)

$$18 - \frac { 3 - 1 } { ( 2 - i ) ( 5 + 20 i ) }$$

Evaluate

$$\frac{7638}{425}+\frac{14}{425}i\approx 17.971764706+0.032941176i$$

Solution Steps

Subtract 1 from 3 to get 2.

$$18-\frac{2}{\left(2-i\right)\left(5+20i\right)}$$

Multiply complex numbers 2-i and 5+20i like you multiply binomials.

$$18-\frac{2}{2\times 5+2\times \left(20i\right)-i\times 5-20i^{2}}$$

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

$$18-\frac{2}{2\times 5+2\times \left(20i\right)-i\times 5-20\left(-1\right)}$$

Do the multiplications in 2\times 5+2\times \left(20i\right)-i\times 5-20\left(-1\right).

$$18-\frac{2}{10+40i-5i+20}$$

Combine the real and imaginary parts in 10+40i-5i+20.

$$18-\frac{2}{10+20+\left(40-5\right)i}$$

Do the additions in 10+20+\left(40-5\right)i.

$$18-\frac{2}{30+35i}$$

Multiply both numerator and denominator of \frac{2}{30+35i} by the complex conjugate of the denominator, 30-35i.

$$18-\frac{2\left(30-35i\right)}{\left(30+35i\right)\left(30-35i\right)}$$

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

$$18-\frac{2\left(30-35i\right)}{30^{2}-35^{2}i^{2}}$$

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

$$18-\frac{2\left(30-35i\right)}{2125}$$

Multiply 2 times 30-35i.

$$18-\frac{2\times 30+2\times \left(-35i\right)}{2125}$$

Do the multiplications in 2\times 30+2\times \left(-35i\right).

$$18-\frac{60-70i}{2125}$$

Divide 60-70i by 2125 to get \frac{12}{425}-\frac{14}{425}i.

$$18+\left(-\frac{12}{425}+\frac{14}{425}i\right)$$

Combine the real and imaginary parts in numbers 18 and -\frac{12}{425}+\frac{14}{425}i.

$$18-\frac{12}{425}+\frac{14}{425}i$$

Add 18 to -\frac{12}{425}.

$$\frac{7638}{425}+\frac{14}{425}i$$

Real Part

$$\frac{7638}{425} = 17\frac{413}{425} = 17.971764705882354$$

Solution Steps

Subtract 1 from 3 to get 2.

$$Re(18-\frac{2}{\left(2-i\right)\left(5+20i\right)})$$

Multiply complex numbers 2-i and 5+20i like you multiply binomials.

$$Re(18-\frac{2}{2\times 5+2\times \left(20i\right)-i\times 5-20i^{2}})$$

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

$$Re(18-\frac{2}{2\times 5+2\times \left(20i\right)-i\times 5-20\left(-1\right)})$$

Do the multiplications in 2\times 5+2\times \left(20i\right)-i\times 5-20\left(-1\right).

$$Re(18-\frac{2}{10+40i-5i+20})$$

Combine the real and imaginary parts in 10+40i-5i+20.

$$Re(18-\frac{2}{10+20+\left(40-5\right)i})$$

Do the additions in 10+20+\left(40-5\right)i.

$$Re(18-\frac{2}{30+35i})$$

Multiply both numerator and denominator of \frac{2}{30+35i} by the complex conjugate of the denominator, 30-35i.

$$Re(18-\frac{2\left(30-35i\right)}{\left(30+35i\right)\left(30-35i\right)})$$

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

$$Re(18-\frac{2\left(30-35i\right)}{30^{2}-35^{2}i^{2}})$$

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

$$Re(18-\frac{2\left(30-35i\right)}{2125})$$

Multiply 2 times 30-35i.

$$Re(18-\frac{2\times 30+2\times \left(-35i\right)}{2125})$$

Do the multiplications in 2\times 30+2\times \left(-35i\right).

$$Re(18-\frac{60-70i}{2125})$$

Divide 60-70i by 2125 to get \frac{12}{425}-\frac{14}{425}i.

$$Re(18+\left(-\frac{12}{425}+\frac{14}{425}i\right))$$

Combine the real and imaginary parts in numbers 18 and -\frac{12}{425}+\frac{14}{425}i.

$$Re(18-\frac{12}{425}+\frac{14}{425}i)$$

Add 18 to -\frac{12}{425}.

$$Re(\frac{7638}{425}+\frac{14}{425}i)$$

The real part of \frac{7638}{425}+\frac{14}{425}i is \frac{7638}{425}.

$$\frac{7638}{425}$$