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Algebra II : Imaginary Numbers

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #111 : Imaginary Numbers

$Simplify:\ (3-5i)-(2+7i)$

Possible Answers:

41+11i

-41-11i

1-12i

1+2i

Correct answer:

1-12i

Explanation:

When combining imaginary numbers we must first distribute the negative sign.

(3-5i)-(2+7i)=3-5i-2-7i

Now combine like terms.

3-2-5i-7i=1-12i

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Example Question #44 : Basic Operations With Complex Numbers

Simplify the term: $i^{20}-i^{25}$

Possible Answers:

1-i

i+1

-1-i

-2i

Correct answer:

1-i

Explanation:

Write the value of the imaginary value.

$i=\sqrt{-1}$

Determine the values of the first several powers of .

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

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

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

$i^5 = i\cdot i^4= i$

Notice that this pattern will repeat.

Rewrite the expression in terms of the product of powers.

$i^{20}-i^{25} =( i^4)^5- i(i^{24}) = (1)^5-i((i^4)^6)$

Simplify i((i^4)^6) .

i((i^4)^6) = i((1)^6)=i

Replace the term.

(1)^5-i((i^4)^6) = 1-i

The answer is: 1-i

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Example Question #2065 : Mathematical Relationships And Basic Graphs

Simplify: i(3-11i)

Possible Answers:

$3i-11i^{2}$

3-11i

11+3i

11-3i

Correct answer:

11+3i

Explanation:

Simplify: i(3-11i)

$First\ distribute\ the\ i--> i(3-11i)=3i-11i^{2}$

$We\ must\ remember\ that\ i^{2}=-1$

$Substitute\ -1\ for\ i^{2}\ to\ get:$

3i-(11)(-1)=3i+11=11+3i

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Example Question #2066 : Mathematical Relationships And Basic Graphs

$Simplify: (2-5i)^{2}$

Possible Answers:

$4-25i^{2}$

$4+25i^{2}$

21-20i

Correct answer:

21-20i

Explanation:

$When\ a\ binomial\ is\ squared\ we\ must\ multiply\ it\ by\ itself.$

$(2-5i)^{2}=(2-5i)(2-5i)$

$Double\ distribute\ or\ FOIL\ to\ get:(2-5i)(2-5i)=4-10i-10i+25i^{2}$

$4-10i-10i+25i^{2}=4-20i+25(i^{2})$

$substitute\ (i^2=-1)$

4-20i+25(-1)=4-20i-25=-21-20i

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Example Question #112 : Imaginary Numbers

Evaluate: $-2(i^{10}-i^{25})$

Possible Answers:

2+2i

2-2i

2-i

-2-2i

Correct answer:

2+2i

Explanation:

The value of $i=\sqrt{-1}$ . We can rewrite the expression as the product of exponents to simplify.

$-2(i^{10}-i^{25})=-2((i^{2})^5-(i^{5})^{5})$

Identify the values of i^2 and i^5 .

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

$i^5 = i^2 \cdot i^2 \cdot i = (-1)(-1)i = i$

Replace the values and simplify.

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

The answer is: 2+2i

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Example Question #2071 : Mathematical Relationships And Basic Graphs

Solve: (i^2+2i^8)^2

Possible Answers:

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

Correct answer:

Explanation:

Evaluate the inner terms of the imaginary values.

The value of $i=\sqrt{-1}$ .

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

i^8 = (i^2)^4 = (-1)^4 =1

The expression becomes:

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

The answer is:

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Example Question #114 : Imaginary Numbers

Express the following number in the form a+bi , where and are real numbers:

$\frac{17}{4-3i}$

Possible Answers:

$17+\frac{51}{4}i$

$\frac{68}{25}+\frac{51}{25}i$

$\frac{68}{25}-\frac{51}{25}i$

$\frac{68}{7}-\frac{51}{7}i$

Correct answer:

$\frac{68}{25}+\frac{51}{25}i$

Explanation:

Since there is a complex number in the form a+bi in the denominator of the given expression, multiply the numerator and the denominator of the expression by its complex conjugate:

$\frac{17}{4-3i}=(\frac{17}{4-3i})(\frac{4+3i}{4+3i})$

$=\frac{17(4+3i)}{(4-3i)(4+3i)}$

$=\frac{68+51i}{16-9i^2}$

$=\frac{68+51i}{16+9}$

$=\frac{68+51i}{25}$

$=\frac{68}{25}+\frac{51}{25}i$

This number is now expressed in the form a+bi , where $a=\frac{68}{25}$ and $b=\frac{51}{25}$ . Hence, the correct answer is $\frac{68}{25}+\frac{51}{25}i$ .

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Example Question #4731 : Algebra Ii

Evaluate: $2i^{243}-i^3$

Possible Answers:

-2i

-3i

Correct answer:

Explanation:

Recall that $i=\sqrt{-1}$ .

$i^3 = i^2\cdot i =(-1)i = -i$

Rewrite the power of 243 as a product of a power of three.

$2i^{243}-i^3 =2(i^{3})^{81}-i^3 =2(-i)^{81}-(-i)$

Simplify the expression.

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

The answer is:

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Example Question #4732 : Algebra Ii

Determine the value of: $3i^{1000}$

Possible Answers:

-3i

Correct answer:

Explanation:

In order to determine the exact value of $i^{1000}$ , we will need to rewrite this exponent as a product of exponents.

$i^{1000} =( i^4)^{250}$

Recall that $i=\sqrt{-1}$ .

i^2=-1

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

$( i^4)^{250}= (1)^{250} = 1$

Therefore,

$3i^{1000} = 3(1) =3$

The answer is:

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Example Question #2071 : Mathematical Relationships And Basic Graphs

Add: $4i^4-8i^{20}-12i^2$

Possible Answers:

$\textup{The terms are simplified as is.}$

Correct answer:

Explanation:

The expression contain imaginary terms. The values can be simplified.

Recall that:

$i=\sqrt{-1}$

This indicates that:

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

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

Rewrite the expression. For powers higher than the given terms, we can rewrite that power as a product of exponents.

$4i^4-8i^{20}-12i^2 = 4(1)-8(i^4)^5-12(-1)$

4-8(1)^5+12 = 4-8+12 = 8

The answer is:

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Algebra II : Imaginary Numbers

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #111 : Imaginary Numbers

Example Question #44 : Basic Operations With Complex Numbers

Example Question #2065 : Mathematical Relationships And Basic Graphs

Example Question #2066 : Mathematical Relationships And Basic Graphs

Example Question #112 : Imaginary Numbers

Example Question #2071 : Mathematical Relationships And Basic Graphs

Example Question #114 : Imaginary Numbers

Example Question #4731 : Algebra Ii

Example Question #4732 : Algebra Ii

Example Question #2071 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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