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Algebra II : Complex Imaginary Numbers

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Example Question #41 : Imaginary Numbers

Evaluate: $\frac{2i}{i-3}+\frac{i}{i+1}$

Possible Answers:

$-\frac{1}{10}i+\frac{17}{10}$

$-\frac{7}{10}i+\frac{7}{10}$

$-\frac{3}{20}i-\frac{7}{20}$

$-\frac{1}{20}i+\frac{7}{20}$

$-\frac{1}{10}i+\frac{7}{10}$

Correct answer:

$-\frac{1}{10}i+\frac{7}{10}$

Explanation:

Identify the least common denominator of the imaginary terms by using the FOIL method.

(i-3)(i+1) = (i)(i)+(i)(1)+(-3)(i)+(-3)(1)

Simplify the terms. Recall that:

$i=\sqrt{-1}$ and $i^2= i\cdot i = -1$

The expression becomes:

i^2-2i-3 = (-1)-2i-3 = -2i-4

Convert the fractions of the original problem.

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

Simplify the numerators.

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

Replace the i^2 term.

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

Replace the denominator with the value of the LCD.

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

Factor out a negative one.

$\frac{-i-3}{-2i-4} = \frac{-1(i+3)}{-1(2i+4)} = \frac{i+3}{2i+4}$

Simplify by multiplying both the top and bottom by the conjugate of the denominator. Factor both the top and bottom using the FOIL method.

$\frac{i+3}{2i+4}\cdot \frac{2i-4}{2i-4} = \frac{2i^2+2i-12}{4i^2-16}$

Replace the i^2 term.

$\frac{2i^2+2i-12}{4i^2-16}= \frac{2(-1)+2i-12}{4(-1)-16}= \frac{2i-14}{-20}$

Simplify the fraction.

The answer is: $-\frac{1}{10}i+\frac{7}{10}$

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Example Question #42 : Imaginary Numbers

Simplify: $\frac{2i-3}{i-3}$

Possible Answers:

$\frac{3}{10}i-\frac{11}{10}$

$-\frac{3}{10}i+\frac{11}{10}$

$-\frac{3}{5}i+\frac{11}{5}$

$-\frac{5}{8}i+\frac{7}{8}$

$-\frac{3}{8}i+\frac{5}{8}$

Correct answer:

$-\frac{3}{10}i+\frac{11}{10}$

Explanation:

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

$\frac{2i-3}{i-3} \cdot \frac{i+3}{i+3}$

Since the i-terms are imaginary numbers, write out the actual value of and i^2 .

$i=\sqrt{-1}$

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

Simplify both top and bottom by the FOIL method.

(2i-3)(i+3) = 2i^2+3i-9 = 2(-1)+3i-9 = 3i-11

(i-3)(i+3) = i^2-9 = (-1)-9 = -10

Divide the two terms.

$\frac{ 3i-11}{-10}$

The answer is: $-\frac{3}{10}i+\frac{11}{10}$

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Example Question #43 : Imaginary Numbers

Simplify: $-\frac{i}{i-1}$

Possible Answers:

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

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

$\frac{1}{4}i-\frac{1}{4}$

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

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

Correct answer:

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

Explanation:

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

$-\frac{i}{i-1} \cdot \frac{i+1}{i+1}$

Simplify the top and bottom. The bottom can be simplified by the FOIL method.

i(i+1) = i^2+i

(i-1)(i+1) = i^2-1

The expression becomes: $-\frac{i^2+i}{i^2-1}$

Recall that $i=\sqrt{-1}$ and $i^2= i\cdot i = -1$ . Replace the values.

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

The answer is: $\frac{1}{2}i-\frac{1}{2}$

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Example Question #44 : Imaginary Numbers

Simplify, if possible: $\frac{1-i}{2-i}$

Possible Answers:

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

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

$-\frac{1}{10}i+\frac{3}{10}$

$-\frac{3}{5}i+\frac{3}{5}$

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

Correct answer:

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

Explanation:

In order to simplify this, we will need to multiply both the top and bottom by the conjugate of the denominator.

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

Simplify the top and bottom by FOIL method. Recall that: $i=\sqrt{-1}$ and $i^2= i\cdot i = -1$ .

Start with the numerator.

(1-i)(2+i) = 2-2i+i-i^2 = -i^2-i+2

-i^2-i+2 = -(-1)-i+2 = 1-i+2 = 3-i

Solve the denominator.

(2-i)(2+i) = 4-i^2 = 4-(-1) =5

Divide the numerator and denominator.

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

The answer is: $-\frac{1}{5}i+\frac{3}{5}$

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Example Question #45 : Imaginary Numbers

Simplify: $\frac{i^3+1}{i^{9}-2}$

Possible Answers:

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

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

$\frac{1}{5}i-\frac{4}{5}$

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

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

Correct answer:

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

Explanation:

Write out the power terms of .

$i=\sqrt{-1}$

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

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

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

Then,

$i^9 = (i^4)^2\cdot i =(1)^2 \cdot i =i$

Replace the terms.

$\frac{i^3+1}{i^{9}-2} = \frac{1-i}{i-2}$

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

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

Simplify both terms by the FOIL method.

(1-i)(i+2) = (1)(i)+(1)(2)+(-i)(i)+(-i)(2)

i+2-i^2-2i = i+2-(-1)-2i =3-i

(i-2)(i+2) = i^2-4 = (-1)-4 = -5

Divide the numerator by the denominator.

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

The answer is: $\frac{1}{5}i-\frac{3}{5}$

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Example Question #46 : Imaginary Numbers

Evaluate: $\frac{i^2-i-3}{i}$

Possible Answers:

1-4i

4i-1

2i-1

-3i

2i-2

Correct answer:

4i-1

Explanation:

Simplify the numerator. Recall that $i=\sqrt{-1}$ and $i^2= i\cdot i = -1$ .

Replace the value of i^2 .

i^2-i-3 = -1-i-3 = -i-4

Rewrite the fraction and split it into two terms.

$\frac{-i-4}{i} = -\frac{i}{i}-\frac{4}{i}$

Simplify both terms.

$-1-\frac{4}{i}$

To simplify $\frac{4}{i}$ , multiply on the top and bottom.

$\frac{4}{i}\cdot \frac{i}{i} = \frac{4i}{i^2} = \frac{4i}{-1} = -4i$

This means that: $-1-\frac{4}{i}=-1-(-4i) = 4i-1$

The answer is: 4i-1

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Example Question #47 : Imaginary Numbers

Solve: $\frac{6}{4i-3}$

Possible Answers:

$-\frac{4}{5}i-\frac{9}{5}$

$\frac{9}{10}i+\frac{7}{10}$

$-\frac{18}{25}i-\frac{24}{25}$

$-\frac{9}{10}i-\frac{7}{10}$

$-\frac{24}{25}i-\frac{18}{25}$

Correct answer:

$-\frac{24}{25}i-\frac{18}{25}$

Explanation:

Multiply the numerator and denominator by the conjugate of the denominator.

$\frac{6}{4i-3} \cdot \frac{4i+3}{4i+3}$

Simplify the top and bottom.

6(4i+3) = 24i+18

(4i-3)(4i+3) = 16i^2-9

The fraction becomes: $\frac{24i+18}{16i^2-9}$

Recall that i^2 =-1 since $i=\sqrt{-1}$ . Replace the value.

$\frac{24i+18}{16i^2-9}=\frac{24i+18}{16(-1)-9} = \frac{24i+18}{-25}$

The answer is: $-\frac{24}{25}i-\frac{18}{25}$

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Example Question #41 : Imaginary Numbers

(2+i)(4+3i)

Possible Answers:

-5+10i

5-10i

5+10i

None of these.

-5-10i

Correct answer:

5+10i

Explanation:

(2+i)(4+3i)

FOIL:

8+6i+4i+3i^2

Simplify like terms:

8+10i+3i^2

Simplify i^2 :

8+10i+3(-1)

$\mathbf{5+10i}$

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Example Question #42 : Imaginary Numbers

Simplify: $\frac{2}{(2i+1)(2i-1)}$

Possible Answers:

$\frac{8}{25}i-\frac{6}{25}$

$\textup{The answer is not given.}$

$-\frac{8}{25}i-\frac{6}{25}$

$-\frac{2}{5}$

$-\frac{2}{3}$

Correct answer:

$-\frac{2}{5}$

Explanation:

Use the FOIL method to simplify the denominator.

(2i+1)(2i-1) = (2i)(2i)+(2i)(-1)+(1)(2i)+(1)(-1)

= 4i^2-1

The imaginary term is defined as: $i=\sqrt{-1}$

This means that: $i^2= i\cdot i = -1$

Replace the term.

4i^2-1 = 4(-1)-1=-5

The answer is: $-\frac{2}{5}$

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Example Question #50 : Imaginary Numbers

Simplify: 7i^3*6i^2

Possible Answers:

42i

42i^5

-42i

Correct answer:

42i

Explanation:

We know that $i=\sqrt{-1}$ and i^2=-1 . Therefore 7i^3*6i^2=42i^5 . We know i^5=i^2*i^2*i . Therefore we will have just . Our final answer is 42*i=42i .

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Algebra II : Complex Imaginary Numbers

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Imaginary Numbers

Example Question #42 : Imaginary Numbers

Example Question #43 : Imaginary Numbers

Example Question #44 : Imaginary Numbers

Example Question #45 : Imaginary Numbers

Example Question #46 : Imaginary Numbers

Example Question #47 : Imaginary Numbers

Example Question #41 : Imaginary Numbers

Example Question #42 : Imaginary Numbers

Example Question #50 : Imaginary Numbers

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