Complex Imaginary Numbers

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Algebra II › Complex Imaginary Numbers

Questions 1 - 10

1

Which of the following is equal to $-\sqrt{-1}$ ?

1

Explanation

Each of the following are true:

$i = \sqrt{-1}$

$i^2 = (\sqrt{-1})^2 = -1$

$i^3 = (\sqrt{-1})^3 = -\sqrt{-1}$

$i^4 = (\sqrt{-1})^4 = 1$

Therefore, the correct answer is .

2

Evaluate: $\frac{i^6+1}{i-1}$

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

Explanation

The imaginary term is equivalent to $\sqrt{-1}$ .

This means that:

Substitute this term back into the numerator.

$\frac{i^6+1}{i-1}=\frac{-1+1}{i-1} = \frac{0}{i-1} =0$

There is no need to use extra steps such as multiplying by the conjugate of the denominator to simplify.

The answer is:

3

Simplify: $\frac{i^{60}-1}{i^3-1}$

$\textup{Undefined.}$

Explanation

Write the first few terms of the imaginary term.

$i=\sqrt{-1}$

$i^2 = i\cdot i =-1$

$i^3 = i^2\cdot i = -i$

$i^4 = i^2\cdot i^2=1$

Notice that these terms will be in a pattern for higher order imaginary terms.

Rewrite the numerator using the product of exponents.

$\frac{i^{60}-1}{i^3-1} =\frac{(i^4)^{15}-1}{i^3-1} = \frac{1-1}{-i-1} = 0$

The answer is:

4

Evaluate:

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

Explanation

We can set in the cube of a binomial pattern:

$\left ( A + B\right ) ^{3} = A ^{3} + 3 A^{2 }B + 3AB^{2} + B^{3}$

$= 2 ^{3} + 3 \cdot 2 ^{2 } \cdot 3i + 3 \cdot 2 \cdot (3i)^{2} + (3i)^{3}$

$= 2^{3} + 3 \cdot 2^{2} \cdot 3i + 3 \cdot 2 \cdot 3 ^{2}\cdot i^{2} + 3^{3} \cdot i^{3}$

$= 8 + 36i + 54 i^{2} + 27 i^{3}$

5

Multiply:

Explanation

Use the FOIL technique:

$= 7 \cdot 1 - 7 \cdot 2i + 3i \cdot 1 - 3i \cdot 2i$

$= 7 - 14i + 3i - 6i ^{2}$

6

Evaluate: $\frac{2i}{5-2i}$

$\frac{10i-4}{29}$

$\frac{10i+4}{29}$

$\frac{10i-4}{21}$

$\frac{10i-4}{5}$

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

Explanation

Multiply the top and bottom by the conjugate of the denominator.

$\frac{2i}{5-2i}\cdot \frac{5+2i}{5+2i}$

Distribute the numerator and FOIL the denominator.

$\frac{2i}{5-2i}\cdot \frac{5+2i}{5+2i} = \frac{10i+4i^2}{25-4i^2}$

Recall that $i=\sqrt{-1}$ . This means that $i^2 = i\cdot i = -1$ .

Replace the terms.

$\frac{10i+4(-1)}{25-4(-1)} = \frac{10i-4}{29}$

The answer is: $\frac{10i-4}{29}$

7

Simplify:

Explanation

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

8

Simplify: $\frac{i^{150}-2i^{8}}{3i^{100}-6}$

$\frac{1}{9}$

$\textup{The expression cannot be factored.}$

Explanation

Write out some of the imaginary terms.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

The powers of the imaginary number can be rewritten using the product of exponents.

$i^{150} = (i^2)^{75} = (-1)^{75} = -1$

$i^{100} = (i^4)^{25} = (1)^{25} = 1$

Replace all the terms back into the expression.

$\frac{i^{150}-2i^{8}}{3i^{100}-6} = \frac{-1-2(1)}{3(1)-6} = \frac{-3}{-3} = 1$

The answer is:

9

Simplify:

$\frac{8-3i}{2-i}$

$\frac{19+2i}{5}$

$\frac{7+2i}{3}$

Explanation

To get rid of the fraction, multiply the numerator and denominator by the conjugate of the denominator.

$\frac{8-3i}{2-i}\left(\frac{2+i}{2+i}\right)$

Now, multiply and simplify.

$\frac{8-3i}{2-i}\left(\frac{2+i}{2+i}\right)=\frac{16+8i-6i-3i^2}{4-i^2}$

Remember that

$\frac{16+8i-6i-3i^2}{4-i^2}=\frac{16+2i-3(-1)}{4-(-1)}=\frac{19+2i}{5}$

10

Simplify: $\frac{3i}{i+9}$

$\frac{27}{82}i+\frac{3}{82}$

$\frac{27}{35}i+\frac{3}{35}$

$\frac{27}{82}i-\frac{3}{82}$

$\frac{27}{13}i+\frac{3}{13}$

$-3i+\frac{1}{9}$

Explanation

In order to simplify this expression, we will need to multiply the numerator and denominator by the conjugate of the denominator.

$\frac{3i}{i+9}\cdot \frac{i-9}{i-9}$

Simplify both the top and bottom. Recall that: $i=\sqrt{-1}$

Divide the numerator with the denominator.

$\frac{ -27i-3}{-82}$

Simplify this term and split the fraction.

The answer is: $\frac{27}{82}i+\frac{3}{82}$

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