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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

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All SAT II Math II Resources

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Example Questions

Example Question #21 : Sat Subject Test In Math Ii

$\overline{x}$ denotes the complex conjugate of .

If $x = 5 + i \sqrt{2}$ , then evaluate $x^{4} \cdot \overline{x} ^{4}$ .

Possible Answers:

279,841

390,609

2,500

390,641

531, 441

Correct answer:

531, 441

Explanation:

Applying the Power of a Product Rule:

$x^{4} \cdot \overline{x} ^{4}=(x \overline{x})^{4}$

The complex conjugate of an imaginary number x = a+ bi is $\overline{x} = a - bi$ ; the product of the two is

$x \overline{x} = a^{2} + b^{2}$

$x = 5 + i \sqrt{2}$ , so, setting $a = 5 , b = \sqrt{2}$ in the above pattern:

$x \overline{x} = 5^{2} + (\sqrt{2})^{2} = 25 + 2= 27$

Consequently,

$x^{4} \cdot \overline{x} ^{4}=(x \overline{x})^{4} = 27 ^{4} = 531, 441$

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Example Question #11 : Real And Complex Numbers

Let and be complex numbers. $\overline{x}$ and $\overline{y}$ denote their complex conjugates.

$\overline{x} \cdot \overline{y } = 7 + 8 i \sqrt{2}$

$y = 8 - i\sqrt{3}$

Evaluate .

Possible Answers:

$-7 + 8 i \sqrt{2}$

$-7 - 8 i \sqrt{2}$

$7 - 8 i \sqrt{2}$

$8 \sqrt{2} - 7i$

$8 \sqrt{2}+ 7i$

Correct answer:

$7 - 8 i \sqrt{2}$

Explanation:

Knowing the actual values of and is not necessary to solve this problem. The product of the complex conjugates of two numbers is equal to the complex conjugate of the product of the numbers; that is,

$\overline{xy }= \overline{x} \cdot \overline{y }$

We are given that $\overline{xy }= \overline{x} \cdot \overline{y } = 7 + 8 i \sqrt{2}$ . is therefore the conjugate of $\overline{xy }$ , or $7 - 8 i \sqrt{2}$ .

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Example Question #21 : Number Theory

$\overline{x}$ denotes the complex conjugate of .

If $x = \frac{1}{10} - \frac{1}{5} i$ , then evaluate $\overline{x}^{2} - x^{2}$ .

Possible Answers:

$\frac{2}{25} i$

$-\frac{2}{25}$

$\frac{4}{25 }i$

$\frac{1}{25}$

$-\frac{4}{25 }$

Correct answer:

$\frac{2}{25} i$

Explanation:

By the difference of squares pattern,

$\overline{x}^{2} - x^{2} = (\overline{x}+ x) ( \overline{x} - x)$

If $x = \frac{1}{10} - \frac{1}{5} i$ , then $\overline{x} = \frac{1}{10} + \frac{1}{5} i$ .

Consequently:

$\overline{x} + x =\left ( \frac{1}{10} + \frac{1}{5} i \right ) + \left ( \frac{1}{10} - \frac{1}{5} i \right )$

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

$= \frac{1}{5}$

$\overline{x} - x =\left ( \frac{1}{10} + \frac{1}{5} i \right ) - \left ( \frac{1}{10} - \frac{1}{5} i \right )$

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

$= \frac{2}{5} i$

Therefore,

$\overline{x}^{2} - x^{2} = (\overline{x}+ x) ( \overline{x} - x) = \frac{1}{5} \cdot \frac{2}{5} i = \frac{2}{25} i$

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Example Question #21 : Sat Subject Test In Math Ii

The fraction $\frac{1+i}{1-i}$ is equivalent to which of the following?

Possible Answers:

$\frac{1}{i}$

Undefined

Correct answer:

Explanation:

Start by multiplying the fraction by $\frac{1+i}{1+i}$ .

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

Since i^2=-1 , we can then simplify the fraction:

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

Thus, the fraction is equivalent to .

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Example Question #22 : Sat Subject Test In Math Ii

The fraction $\frac{5+3i}{2-2i}$ is equivalent to which of the following?

Possible Answers:

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

1+2i

$4-\frac{1}{2}i$

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

Correct answer:

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

Explanation:

Start by multiplying both the denominator and the numerator by the conjugate of 2-2i , which is 2+2i .

$\frac{5+3i}{2-2i}(\frac{2+2i}{2+2i})=\frac{10+10i+6i+6i^2}{4-4i^2}$

Next, recall i^2=-1 , and combine like terms.

$\frac{10+10i+6i+6i^2}{4-4i^2}=\frac{10+16i+6(-1)}{4-4(-1)}=\frac{4+16i}{8}$

Finally, simplify the fraction.

$\frac{4+16i}{8}=\frac{1}{2}+2i$

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Example Question #23 : Sat Subject Test In Math Ii

Evaluate $i^{99}$

Possible Answers:

99i

-99i

Correct answer:

Explanation:

To evaluate a power of , divide the exponent by 4 and note the remainder.

$99 \div 4 = 24\textrm{ R }3$

The remainder is 3, so

$i^{99} = i^{3}$

Consequently, using the Product of Powers Rule:

$i^{99} = i^{2} \cdot i = -1 \cdot i = -i$

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Example Question #21 : Number Theory

Let be a complex number. $\overline{x}$ denotes the complex conjugate of .

$x+\overline{ x} = 20$ and $x \overline{ x} = 200$ .

How many of the following expressions could be equal to ?

(a) 10 + 10i

(b) 10 - 10i

(d) -10 - 10i

Possible Answers:

One

Four

Two

Three

None

Correct answer:

Two

Explanation:

is a complex number, so x = a+ bi for some real a ,b ; also, $\overline{x} = a-bi$ .

Therefore,

$x+\overline{ x} = (a+ bi) + (a-bi) = a+ a + bi - bi = 2a$

Substituting:

$x+\overline{ x} = 20$

2a = 20

a = 10

Therefore, we can eliminate choices (c) and (d).

Also, the product

$x \overline{ x} = (a+bi)(a-bi) = a^{2} +b^{2}$

Setting $x \overline{ x} = 200$ and substituting 10 for , we get

$10^{2} +b^{2} = 200$

$100 +b^{2} = 200$

$b^{2} = 100$

$b = \pm 10$

Therefore, either x = 10 + 10i or x = 10 - 10i - making two the correct response.

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Example Question #1 : Mathematical Relationships

Add the following numbers: 137+189+774

Possible Answers:

1180

1290

1200

1190

1100

Correct answer:

1100

Explanation:

Add the ones digits.

7+9+4 = 20

Add the tens digits with the two as the carryover.

3+8+7+(2) = 20

Add the hundreds digits with the two as the carryover.

1+1+7+(2) = 11

Combine this number with the ones digits of the previous calculations.

The answer is: 1100

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Example Question #1 : Elementary Operations

Add the numbers: 256+781+64

Possible Answers:

1101

1011

901

1111

1901

Correct answer:

1101

Explanation:

Add the ones digits.

6+1+4 = 11

Carry the one, and add the tens digits.

5+8+6+(1) = 20

Carry the two, and add the hundreds digits.

2+7+0+(2) = 11

Combine this number with the ones digit of the other numbers.

The answer is: 1101

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Example Question #2 : Elementary Operations

Add the numbers: 34+497+89

Possible Answers:

520

550

510

620

650

Correct answer:

620

Explanation:

Add the ones digits.

4+7+9 = 20

Carry the two, and add the tens digits.

3+9+8+(2) = 22

Carry the two, and add the hundreds places.

0+4+0+(2) = 6

Combine all ones digits.

The answer is: 620

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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #21 : Sat Subject Test In Math Ii

Example Question #11 : Real And Complex Numbers

Example Question #21 : Number Theory

Example Question #21 : Sat Subject Test In Math Ii

Example Question #22 : Sat Subject Test In Math Ii

Example Question #23 : Sat Subject Test In Math Ii

Example Question #21 : Number Theory

Example Question #1 : Mathematical Relationships

Example Question #1 : Elementary Operations

Example Question #2 : Elementary Operations

All SAT II Math II Resources

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