Solve (3-i)(2-(1i)/(2+ui)(2-ui)

$$\frac { ( 3 - i ) ( 2 - ( 1 i ) } { ( 2 + u i ) ( 2 - u i ) }$$

Evaluate

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

Solution Steps

Multiply complex numbers 3-i and 2-i like you multiply binomials.

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

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

$$\frac{3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right)}{\left(2+ui\right)\left(2-ui\right)}$$

Do the multiplications in 3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right).

$$\frac{6-3i-2i-1}{\left(2+ui\right)\left(2-ui\right)}$$

Combine the real and imaginary parts in 6-3i-2i-1.

$$\frac{6-1+\left(-3-2\right)i}{\left(2+ui\right)\left(2-ui\right)}$$

Do the additions in 6-1+\left(-3-2\right)i.

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

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

$$\frac{5-5i}{2^{2}-\left(ui\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

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

Expand \left(ui\right)^{2}.

$$\frac{5-5i}{4-u^{2}i^{2}}$$

Calculate i to the power of 2 and get -1.

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

Multiply -1 and -1 to get 1.

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

Expand

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

Solution Steps

Multiply complex numbers 3-i and 2-i like you multiply binomials.

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

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

$$\frac{3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right)}{\left(2+ui\right)\left(2-ui\right)}$$

Do the multiplications in 3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right).

$$\frac{6-3i-2i-1}{\left(2+ui\right)\left(2-ui\right)}$$

Combine the real and imaginary parts in 6-3i-2i-1.

$$\frac{6-1+\left(-3-2\right)i}{\left(2+ui\right)\left(2-ui\right)}$$

Do the additions in 6-1+\left(-3-2\right)i.

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

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

$$\frac{5-5i}{2^{2}-\left(ui\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

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

Expand \left(ui\right)^{2}.

$$\frac{5-5i}{4-u^{2}i^{2}}$$

Calculate i to the power of 2 and get -1.

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

Multiply -1 and -1 to get 1.

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