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GRE Subject Test: Math : Operations on Complex Numbers

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

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Example Questions

Example Question #1 : Operations On Complex Numbers

Expand and Simplify: (4-3i)(2+3i)(2-3i)

Possible Answers:

39+52i

-20-15i

52-39i

52+39i

Correct answer:

52-39i

Explanation:

Step 1: We will multiply the two complex conjugates: 2x-3 and 2x+3 .

(2-3i)(2+3i)=4+6i-6i-9i^2=4-9i^2

Step 2: Replace i^2 with .

4-9(-1)

Simplify:

4+9(1)
=13

Step 3: Multiply the result of the complex conjugates to the other parentheses, (4-3i) .

13(4-3i)=52-39i

The final answer after the product of all three binomials is 52-39i

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Example Question #1 : Operations On Complex Numbers

Expand: $(i-1)^{10}$ .

Possible Answers:

-32i

32i

32(1-i)

Correct answer:

-32i

Explanation:

Quick Way:
Step 1: Expand (i-1)^2 .

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

Remember: i^2=-1

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

Step 2: $(1-i)^{10}=(i-1)^{2\cdot 5}$

By this equivalence, I can just raise the answer of (1-i)^2 to the power .

$(i-1)^{10}=(-2i)^{5}=-32i^5$

$i^5 \equiv i^1=i$ . Replace i^5 ..

Final answer: -32i

Long Way:

Math work

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Example Question #1 : Operations On Complex Numbers

Multiply:

(2+3i)(3-4i)

Possible Answers:

6+i

18+i

18-i

-i-18

Correct answer:

18+i

Explanation:

Step 1: FOIL:

Recall, FOIL means to multiply the first terms in both binomials together, the outer terms together, the inner terms together, and finally, the last terms together.

(2+3i)(3-4i)=6-8i+9i-12i^2

Step 2: Simplify:

6+i-12i^2

Step 3: Recall: i^2=-1 . Replace and simplify.

6+i-12(-1)
6+i+12
=18+i

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Example Question #1 : Operations On Complex Numbers

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

Possible Answers:

5+2i

6-35i

41+11i

Correct answer:

5+2i

Explanation:

When adding imaginary numbers, simply add the real parts and the imaginary parts.

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

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Example Question #2 : Operations On Complex Numbers

Simplify: (13-7i)-(10-8i)

Possible Answers:

-74-174i

3+i

23-i

3-15i

Correct answer:

3+i

Explanation:

$When\ subtracting\ binomials\ in\ ( )\; first\ distribute\ the\ negative\ sign.$

(13-7i)-(10-8i)=13-7i-10+8i

$Now\ combine\ like\ terms.$

13-10-7i+8i=3+i

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Example Question #4 : Operations On Complex Numbers

What is the value of (3-4i)(2i+3) ?

Possible Answers:

None of the other answers

-6i+17

-6i+1

-6i-17

6i-17

Correct answer:

-6i+17

Explanation:

Distribute and Multiply:

(3-4i)(2i+3)=6i+9-8i^2-12i

Simplify all terms...

6i+9-8(-1)-12i

=6i+9+8-12i

=-6i+17

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Example Question #5 : Operations On Complex Numbers

What is the value: $i^{24}(i^{3}+i^{-2})$ ?

Possible Answers:

-i-1

1+i

-1-i

-(-i+1)

Correct answer:

-1-i

Explanation:

Step 1: Recall the cycle of imaginary numbers to a random power .

If x=1 , then i^1=i

If x=2 , then i^2=-1

If x=3 , then i^3=-i

If x=4 , then i^4=1

If x=5 , then i^5=i^4(i^1)=1(i)=i

and so on....

The cycle repeats every terms.

For ANY number $x\ge 5$ , you can break down that term into smaller elementary powers of i.

Step 2: Distribute the $i^{24}$ to all terms in the parentheses:

$(i^{24})(i^3)+(i^{24})(i^{-2})$ .

Step 3: Recall the rules for exponents:

$\forall\{a,b,c,d\} \in \mathbb{R}::$

$(a^b)\cdot (a^c)=a^{b+c}$

${a^{-b}}=\frac {1}{a^b}$

$(a^b)\cdot (a^{-c})=a^b \cdot \bigg (\frac {1}{a^c}\bigg)=\frac {a^b}{a^c}=a^{b-c}=a^d$

Step 4: Use the rules to rewrite the expression in Step 2:

$(i^{24})\cdot {i^3}=i^{24+3}=i^{27}$

$(i^{24})(i^{-2})=\frac {i^{24}}{i^2}=i^{24-2}=i^{22}$

Step 5: Simplify the results in Step 4. Use the rules in Step 1.:

$i^{27}=(i^{24}\cdot i^3)=1(-i)=-i$

$i^{22}=(i^{20}\cdot i^{2})=1(-1)=-1$

Step 6: Write the answer in a+bi form, where is the real part and is the imaginary part:

We get -1-i

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Example Question #21 : Imaginary Numbers & Complex Functions

Add: (3-5i)+(2+7i)

Possible Answers:

5+2i

6-35i

1-12i

1+12i

Correct answer:

5+2i

Explanation:

When adding complex numbers, we add the real numbers and add the imaginary numbers.

Add: (3-5i)+(2+7i)

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

=5+2i

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Example Question #22 : Imaginary Numbers & Complex Functions

Subtract: (15+7i)-(12+9i)

Possible Answers:

3-2i

3+16i

3-16i

3-5i

Correct answer:

3-2i

Explanation:

In order to subtract complex numbers, we must first distribute the negative sign to the second complex number.

Subtract: (15+7i)-(12+9i)

15+7i-12-9i

15-12+7i-9i=3-2i

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Example Question #1 : Operations On Complex Numbers

Simplify: 3i(-5+2i)

Possible Answers:

15+6i^2

15-6i

6i^2-15i

-6-15i

Correct answer:

-6-15i

Explanation:

Simplify: 3i(-5+2i)

First we must distribute

3i(-5+2i)=-15i+2i^2

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

-15i+6i^2=-15i+6(-1)

$-15i+6(-1)=-6-15i\ (i\ term\ is\ always\ last)$

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GRE Subject Test: Math : Operations on Complex Numbers

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Operations On Complex Numbers

Example Question #1 : Operations On Complex Numbers

Example Question #1 : Operations On Complex Numbers

Example Question #1 : Operations On Complex Numbers

Example Question #2 : Operations On Complex Numbers

Example Question #4 : Operations On Complex Numbers

Example Question #5 : Operations On Complex Numbers

Example Question #21 : Imaginary Numbers & Complex Functions

Example Question #22 : Imaginary Numbers & Complex Functions

Example Question #1 : Operations On Complex Numbers

All GRE Subject Test: Math Resources

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