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SAT II Math II : Number Theory

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

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Example Questions

Example Question #11 : Real And Complex Numbers

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

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

Possible Answers:

390,641

279,841

2,500

531, 441

390,609

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:

$8 \sqrt{2}+ 7i$

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

$8 \sqrt{2} - 7i$

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

$7 - 8 i \sqrt{2}$

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 #11 : Real And Complex Numbers

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

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

Possible Answers:

$\frac{2}{25} i$

$\frac{1}{25}$

$-\frac{4}{25 }$

$-\frac{2}{25}$

$\frac{4}{25 }i$

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 : Number Theory

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

Possible Answers:

Undefined

$\frac{1}{i}$

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 #24 : Number Theory

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

Possible Answers:

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

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

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

1+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 #25 : Number Theory

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 #26 : 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

None

Four

Two

Three

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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All SAT II Math II Resources

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SAT II Math II : Number Theory

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

All SAT II Math II Resources

Example Questions

Example Question #11 : Real And Complex Numbers

Example Question #11 : Real And Complex Numbers

Example Question #11 : Real And Complex Numbers

Example Question #21 : Number Theory

Example Question #24 : Number Theory

Example Question #25 : Number Theory

Example Question #26 : Number Theory

All SAT II Math II Resources

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