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Algebra II : Complex Imaginary Numbers

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Example Questions

Example Question #31 : Imaginary Numbers

Solve: $\frac{6i}{2i-1}$

Possible Answers:

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

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

$\frac{6}{5}i+\frac{12}{5}$

$\textup{The expression is fully simplified.}$

$-\frac{6}{5}i+\frac{12}{5}$

Correct answer:

$-\frac{6}{5}i+\frac{12}{5}$

Explanation:

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

$\frac{6i}{2i-1}\cdot \frac{2i+1}{2i+1}$

Simplify the top and bottom. The value of $i=\sqrt{-1}$ .

6i(2i+1) = 12i^2+6i = 12(-1)+6i = 6i-12

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

Divide the numerator with the denominator.

$\frac{6i-12}{-5} = -\frac{6}{5}i+\frac{12}{5}$

The answer is: $-\frac{6}{5}i+\frac{12}{5}$

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Example Question #32 : Imaginary Numbers

Simplify: $\frac{i^3}{2i-4}$

Possible Answers:

2i-4

$\frac{1}{5}i-\frac{1}{10}$

$\textup{The question is simplified as is.}$

$\frac{1}{2}i-\frac{1}{8}$

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

Correct answer:

$\frac{1}{5}i-\frac{1}{10}$

Explanation:

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

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

Simplify the top and the bottom.

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

This means that:

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

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

Re-substitute the actual values back into the fraction.

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

Reduce and split this fraction.

The correct answer is: $\frac{1}{5}i-\frac{1}{10}$

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Example Question #31 : Imaginary Numbers

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

Possible Answers:

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

$\frac{1}{5}i-\frac{5}{6}$

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

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

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

Correct answer:

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

Explanation:

The values are imaginary. Recall that:

$i=\sqrt{-1}$

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

Rewrite the numerator.

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

Multiply the top and bottom by the conjugate of i-2 .

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

Rewrite the terms using the value of i^2 .

$\frac{2i^2+4i}{i^2-4}=\frac{2(-1)+4i}{-1-4} =\frac{ -2+4i}{-5}$

The answer is: $-\frac{4}{5}i+\frac{2}{5}$

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Example Question #34 : Imaginary Numbers

Simplify: $\frac{i-6}{2i-1}$

Possible Answers:

$\frac{11}{5}i+\frac{8}{5}$

11i-14

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

-11i+14

$-\frac{13}{5}i-\frac{4}{5}$

Correct answer:

$\frac{11}{5}i+\frac{8}{5}$

Explanation:

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

$\frac{i-6}{2i-1} \times \frac{2i+1}{2i+1}$

Multiply the numerator.

(i-6)(2i+1) = (i)(2i)+(i)(1)+(-6)(2i)+(-6)(1)

2i^2+i-12i-6 = 2i^2-11i-6

The term i^2=-1 since $i=\sqrt{-1}$ . Replace the term and simplify.

= 2(-1)-11i-6 = -11i-8

Multiply the denominator.

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

The expression becomes:

$\frac{i-6}{2i-1} \times \frac{2i+1}{2i+1} = \frac{ -11i-8}{-5}$

Dividing the double negatives will result into a positive value.

The answer is: $\frac{11}{5}i+\frac{8}{5}$

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Example Question #31 : Complex Imaginary Numbers

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

Possible Answers:

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

$-\frac{1}{8}i+\frac{1}{8}$

$-\frac{3}{8}i+\frac{1}{2}$

$-\frac{3}{4}i+\frac{3}{8}$

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

Correct answer:

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

Explanation:

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

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

Multiply the top and bottom by the conjugate of the denominator. Use the FOIL method to simplify the denominator.

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

Replace the values of the known imaginary terms.

$\frac{2i^2+2i}{4i^2-4} =\frac{2(-1)+2i}{4(-1)-4}=\frac{2i-2}{-8}$

Simplify this fraction.

The answer is: $-\frac{1}{4}i+\frac{1}{4}$

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Example Question #36 : Imaginary Numbers

Simplify: $\frac{10}{10+i}$

Possible Answers:

$-\frac{10}{101}i+\frac{100}{101}$

$-\frac{10}{101}i-\frac{100}{101}$

$-\frac{10}{101}i-\frac{101}{100}$

$\frac{1}{10}i-\frac{11}{10}$

$\frac{1}{10}i-\frac{10}{11}$

Correct answer:

$-\frac{10}{101}i+\frac{100}{101}$

Explanation:

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

$\frac{10}{10+i} = \frac{10}{i+10}$

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

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

$\frac{10}{i+10} \times \frac{i-10}{i-10} = \frac{10i-100}{i^2-100}$

The value of $i^2= i\cdot i = -1$ . Replace the term.

$\frac{10i-100}{-1-100} = \frac{10i-100}{-101}$

Reduce this fraction.

The answer is: $-\frac{10}{101}i+\frac{100}{101}$

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Example Question #31 : Complex Imaginary Numbers

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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: $i=\sqrt{-1}$

3i(i-9) = 3i^2-27i = 3(-1)-27i = -27i-3

(i+9)(i-9) = i^2-81 = (-1)-81 =-82

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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Example Question #31 : Imaginary Numbers

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

Possible Answers:

$-\frac{12}{5}i+\frac{6}{5}$

$-\frac{12}{5}i+\frac{3}{5}$

$-\frac{6}{5}i+\frac{6}{5}$

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

$-\frac{12}{5}i-\frac{6}{5}$

Correct answer:

$-\frac{12}{5}i-\frac{6}{5}$

Explanation:

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

$\frac{6i+6}{i-3}\times\frac{i+3}{i+3}$

Simplify the top and bottom using the FOIL method.

(6i+6)(i+3) = 6i^2+18i+6i+18 = 6i^2+24i+18

Note that: $i^2= i\cdot i = -1$

Replace the term. The numerator becomes:

6i^2+24i+18 =6(-1)+24i+18 = 24i+12

Simplify the denominator.

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

Divide the numerator with the denominator.

$\frac{ 24i+12}{-10} = -\frac{12}{5}i-\frac{6}{5}$

The answer is: $-\frac{12}{5}i-\frac{6}{5}$

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Example Question #39 : Imaginary Numbers

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

Possible Answers:

$-\frac{9}{10}i+\frac{3}{10}$

$-\frac{3}{10}i+\frac{1}{9}$

$-\frac{3}{4}i+\frac{3}{5}$

$-\frac{1}{3}i+\frac{9}{10}$

$-\frac{3}{10}i+\frac{9}{10}$

Correct answer:

$-\frac{3}{10}i+\frac{9}{10}$

Explanation:

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

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

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

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

Simplify the denominator by FOIL method.

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

Divide the numerator with the denominator.

$\frac{3i-9}{-10}$

The answer is: $-\frac{3}{10}i+\frac{9}{10}$

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Example Question #31 : Complex Imaginary Numbers

Simplify, if possible: $\frac{5}{i-7}$

Possible Answers:

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

$\textup{The expression is already simplified.}$

$-\frac{1}{35}i-\frac{5}{7}$

10i-7

$-\frac{1}{35}i-\frac{1}{7}$

Correct answer:

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

Explanation:

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

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

Expand the top and bottom.

5(i+7) = 5i+35

(i-7)(i+7) = i^2-49

Recall that:

$i=\sqrt{-1}$

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

Replace the imaginary term for i^2 .

$\frac{5}{i-7} \cdot \frac{i+7}{i+7} = \frac{5i+35}{i^2-49} = \frac{5i+35}{(-1)-49}=\frac{5i+35}{-50}$

Simplify this fraction.

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

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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 #31 : Imaginary Numbers

Example Question #32 : Imaginary Numbers

Example Question #31 : Imaginary Numbers

Example Question #34 : Imaginary Numbers

Example Question #31 : Complex Imaginary Numbers

Example Question #36 : Imaginary Numbers

Example Question #31 : Complex Imaginary Numbers

Example Question #31 : Imaginary Numbers

Example Question #39 : Imaginary Numbers

Example Question #31 : Complex Imaginary Numbers

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