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GRE Subject Test: Math : Imaginary Roots of Negative Numbers

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Example Question #6 :Number Theory

Evaluate:

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

Possible Answers:

-62 + 63 i

-4 6+ 9i

-46 + 63 i

8 + 9i

8 - 27i

Correct answer:

-4 6+ 9i

Explanation:

We can set A = 2, B = 3i in the cube of a binomial pattern:

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

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

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

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

= 8 + 36i + 54 (-1) + 27 (-i)

= 8 + 36i - 54 - 27 i

= -46 + 9i

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Example Question #1 :Complex Imaginary Numbers

Simplify the following product:

{} (5+3i)(-2+i)

Possible Answers:

{} -13+11i

{} -10+3i

{}-7-i

{} -13-i

Correct answer:

{} -13-i

Explanation:

Multiply these complex numbers out in the typical way:

${}(5+3i)(-2+i) = -10+5i-6i+3i^2$

and recall that i^2=-1 by definition. Then, grouping like terms we get

{} (-10-3)+(5i-6i) = -13-i

which is our final answer.

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Example Question #1 :Imaginary Numbers & Complex Functions

Simplify:

(5+8i)(6-2i)

Possible Answers:

16+32i

25+8i

7+16i

46+38i

Correct answer:

46+38i

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.

(5+8i)(6-2i)=30-10i+48i-16i^2

Remember that $i=\sqrt-1$ , so i^2=-1 .

Substitute infor i^2

30-10i+48i-16i^2=30+38i-16(-1)=30+38i+16=46+38i

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Example Question #1 :Imaginary Roots Of Negative Numbers

Simplify:

(2+5i)(-8-i)

Possible Answers:

-18-5i

42+11i

-11-42i

12+6i

Correct answer:

-11-42i

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.

(2+5i)(-8-i)=-16-2i-40i-5i^2

Remember that $i=\sqrt-1$ , so i^2=-1 .

Substitute infor i^2

-16-2i-40i-5i^2=-16-42i-5(-1)=-16+5-42i=-11-42i

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Example Question #1 :Imaginary Numbers & Complex Functions

Simplify:

(2+6i)(5+3i)

Possible Answers:

-36-8i

-8+36i

36-8i

12-8i

Correct answer:

-8+36i

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.

(2+6i)(5+3i)=10+6i+30i+18i^2

Remember that $i=\sqrt-1$ , so i^2=-1 .

Substitute infor i^2

10+6i+30i+18i^2=10+36i+18(-1)=10-18+36i=-8+36i

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Example Question #1 :Equations With Complex Numbers

Solve forand:

$ai^{93}+bi^{35}+ai^{24}-bi^{86}=40+20i$

Possible Answers:

a=15, b=15

a=10, b=30

a=40, b=20

a=20, b=40

a=30, b=10

Correct answer:

a=30, b=10

Explanation:

Remember that

$i = \sqrt{-1}\\ i^2 = \sqrt{-1}^2 = -1\\ i^3 = i^2\cdot i = -1 \cdot i = -i\\ i^4 = i^2 \cdot i^2 = (-1)\cdot(-1) = 1\\ i^5 = i^4\cdot i = 1\cdot i = i = \sqrt{-1}\\ i^6 = i^4\cdot i^2 = 1\cdot i^2 = i^2 =-1$

So the powers ofare cyclic.This means that when we try to figure out the value of an exponent of, we can ignore all the powers that are multiples ofbecause they end up multiplying the end result by, and therefore do nothing.

This means that

$ai^{93} + bi^{35} + ai^{24} - bi^{86}\\ = ai^{92+1} + bi^{32+3} + ai^{24+0} - bi^{84+2}\\ = ai^{92}\cdot i^1 + bi^{32}\cdot i^3 + ai^{24}\cdot i^0 - bi^{84}\cdot i^2\\ =a(i^4)^{23}\cdot i^1 + b(i^4)^8\cdot i^3 + a(i^4)^6\cdot i^0 - b(i^4)^{21}\cdot i^2\\ = a (1^{23})\cdot i + b(1^8)\cdot i^3 + a(1^6) - b(1^{21})\cdot i^2$

Now, remembering the relationships of the exponents of, we can simplify this to:

$ai - bi + a - (-b) = ai-bi+a+b\\ = (a+b) + (a-b)i = 40+20i$

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:

a+b=40

a-b=10

No matter how you solve it, you get the values a=30 , b=10 .

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Example Question #1 :Imaginary Numbers & Complex Functions

Simplify: $\sqrt{-243}$

Possible Answers:

$-9i\sqrt{3}$

$9i\sqrt{3}$

None of the Above

Correct answer:

$9i\sqrt{3}$

Explanation:

Step 1: Split the $\sqrt{-243}$ into $\sqrt{243}\cdot\sqrt{-1}$ .

Step 2: Recall that $\sqrt{-1}=i$ , so let's replace it.

We now have: $\sqrt{243}\cdot i$ .

步骤3:简化 $\sqrt{243}$ . To do this, we look at the number on the inside.

$243=3\cdot3\cdot3\cdot3\cdot3$ .

Step 4: Take the factorization of 243 and take out any pairs of numbers. For any pair of numbers that we find, we only takeof the numbers out.

We have a pair of 3's , so ais outside the radical.
We have another pair of 3's , so one more three is put outside the radical.

We need to multiply everything that we bring outside: 3*3=9

Step 5: Thegoes with the 9...

Step 6: The lastafter taking out pairs gets put back into a square root and is written right after the

It will look something like this: $9i\sqrt{3}$

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Example Question #5 :Imaginary Numbers & Complex Functions

$Simplify: 5\sqrt{12i^{2}}$

Possible Answers:

$10i\sqrt{}3$

$2i\sqrt{3}$

$5\sqrt{-12}$

$5i\sqrt{12}$

Correct answer:

$10i\sqrt{}3$

Explanation:

There are two ways to simplify this problem:

Method 1:

$We\ know\ that\ i^2=-1,\ when\ we\ replace\ that\ we\ get:$

$5\sqrt{-12}$

$=5i\sqrt{12}(take\ the\ i\ out)$

$=5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)$

$=10i{\sqrt{3}}\ reduce$

Method 2:

$We\ know\ that\ the\ square\ root\ of\ anything\ squared\ is\ just\ itself$

$\sqrt{i^2}=i$

$5\sqrt{12i^{2}}=5i\sqrt{12}$

$=5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)$

$=10i{\sqrt{3}}\ reduce$

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Example Question #1 :Imaginary Numbers & Complex Functions

$Simplify: \sqrt{-49}-\sqrt{-16}$

Possible Answers:

-11i

7-4i

4-7i

Correct answer:

Explanation:

$First\ simplify\ each\ radical:$

$\sqrt{-49}-\sqrt{-16}=7i-4i$

$Next\ combine\ like\ terms:$

7i-4i=3i

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Example Question #7 :Imaginary Numbers & Complex Functions

$Simplify: \sqrt{-72}$

Possible Answers:

$6{\sqrt{-2}}$

$-6{\sqrt{2}}$

$-6i\sqrt{2}$

$6i{\sqrt{2}}$

Correct answer:

$6i{\sqrt{2}}$

Explanation:

$When\ reducing\ the\ square\ root\ of\ a\ negative\ number\ -\ we\ must\ first\$

$i=\sqrt{-1}$

$\sqrt{-72}=\sqrt{-1}\sqrt{72}=i\sqrt{72}$

$Once\ we\ have\ taken\ out\ the\ i,\ then\ we\ can\ simply\ break\ down$ $the\ number\ under\ the\ radical\ by\ finding\ the\ largest\ perfect\ square\ that\ will go\ into\ it.$

$36\ will\ go\ into\ 72\ so\ we\ break\ this\ down\ to\ become:$

$i\sqrt{36}\sqrt{2}=6i\sqrt{2}$

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GRE Subject Test: Math : Imaginary Roots of Negative Numbers

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #6 :Number Theory

Example Question #1 :Complex Imaginary Numbers

Example Question #1 :Imaginary Numbers & Complex Functions

Example Question #1 :Imaginary Roots Of Negative Numbers

Example Question #1 :Imaginary Numbers & Complex Functions

Example Question #1 :Equations With Complex Numbers

Example Question #1 :Imaginary Numbers & Complex Functions

Example Question #5 :Imaginary Numbers & Complex Functions

Example Question #1 :Imaginary Numbers & Complex Functions

Example Question #7 :Imaginary Numbers & Complex Functions

All GRE Subject Test: Math Resources

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