SAT Math - Real and Complex Numbers

Which answer choice has the greatest real number value?

$i^4 + i^8 + i^{12} +i^ {16}$

$i^2 + i^8 + i^{10} +i^ {16}$

$i^4 + i^6 + i^{14} +i^ {24}$

$i^{40} + i^2 + i^{13} +i^ {11}$

$i + i^3 + i^{7} +i^ {9}$

Explanation

Recall the definition of and its exponents

$i = \sqrt[]{-1}$

because then

$i^5 = i^4\cdot i =1\cdot i =i$ .

We can generalize this to say

$i^n =i^{n-4}$

Any time is a multiple of 4 then . For any other value of we get a smaller value.

For the correct answer each of the terms equal

$i^4 =i^8 =i^{12} = i^{16} =1$

So:

$i^4 + i^8 + i^{12} +i^ {16}$

Because all the alternative answer choices have 4 terms, and each answer choice has at least one term that is not equal to they must all be less than the correct answer.

Multiply: $\left ( 4+ 7i\right )(4 - 7i)$

None of the other responses is correct.

Explanation

This is the product of a complex number and its complex conjugate. They can be multiplied using the pattern

$\left (A + Bi \right )\left (A - Bi \right ) = A^{2} + B^{2}$

with

$\left (4 + 7i \right )\left (4 - 7i \right ) = 4^{2} + 7^{2} = 16 + 49 = 65$

This is not among the given responses.

Evaluate:

$\left ( 7+3i\right )^{2}$

Explanation

Use the square of a sum pattern

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

where :

$\left ( 7+3i\right )^{2} = 7^{2}+ 2 \cdot 7 \cdot 3i + (3i)^{2}$

$= 7^{2}+ 2 \cdot 7 \cdot 3i + 3^{2}i^{2}$

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

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

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

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

Explanation

Start by multiplying both the denominator and the numerator by the conjugate of , which is .

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

Next, recall , 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.

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

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

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

$\frac{2}{25} i$

$-\frac{2}{25}$

$\frac{1}{25}$

$-\frac{4}{25 }$

$\frac{4}{25 }i$

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,

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

Multiply:

$-3 i \cdot 8i \cdot 5i \cdot 6i$

None of the other responses is correct.

Explanation

$-3 i \cdot 8i \cdot 5i \cdot 6i$

$= \left (-3 \cdot 8 \cdot 5 \cdot 6 \right ) \cdot\left ( i \cdot i \cdot i \cdot i \right )$

$= \left (-720 \right ) \cdot\left ( i ^{4}\right )$

$-720 \cdot 1 = -720$

is a complex number; $\overline{x}$ denotes the complex conjugate of .

$x \cdot \overline{x} = 84$

Which of the following could be the value of ?

Any of the numbers in the other four choices could be equal to .

$9 - i \sqrt{3}$

$8 + 2 i \sqrt{5}$

$7 + i \sqrt{35}$

$6 - 4 i \sqrt{3}$

Explanation

The product of a complex number and its complex conjugate $\overline{x} = a- bi$ is

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

Setting and accordingly for each of the four choices, we want to find the choice for which $x \cdot \overline{x} = a^{2} + b^{2} = 84$ :

$6 - 4 i \sqrt{3}$

$a = 6, b = - 4 \sqrt{3}$

$x \cdot \overline{x} = a^{2} + b^{2} = 6^{2}+ (-4 \sqrt{3})^{2} = 36 + 4 ^{2} (\sqrt{3} )^{2} = 36 + 16 \cdot 3 = 36 + 48 = 84$

$7 + i \sqrt{35}$

$a = 7, b = \sqrt{35}$

$x \cdot \overline{x} = a^{2} + b^{2} = 7^{2} + (\sqrt{35})^{2} = 49 + 35 = 84$

$8 + 2 i \sqrt{5}$

$a = 8 , b = 2 \sqrt{5}$

$x \cdot \overline{x} = a^{2} + b^{2} = 8^{2} + (2 \sqrt{5})^{2} = 64 + 2^{2} \cdot (\sqrt{5})^{2} = 64 + 4 \cdot 5 = 64 + 20 = 84$

$9 - i \sqrt{3}$

$a = 9 , b = - \sqrt{3}$

$x \cdot \overline{x} = a^{2} + b^{2} = 9^{2} + (- \sqrt{3})^{2} = 81 + 3 = 84$

For each given value of , $x \cdot \overline{x} = 84$ .

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)

(b)

(c)

(d)

Two

One

None

Three

Four

Explanation

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

Therefore,

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

Substituting:

$x+\overline{ x} = 20$

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 or - making two the correct response.

Which of the following is equal to $i^{77}$ ?

Explanation

To raise to a power, divide the exponent by 4 and note the remainder.

$77 \div 4 = 19\textrm{ R }1$

Raise to the power of that remainder:

$i^{77} = i^{1} = i$

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

$x+\overline{ x} = 16$ and $x-\overline{ x} = 40i$ .

Evaluate .

None of these

Explanation

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

Therefore,

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

Substituting:

$x+\overline{ x} = 16$

Also,

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

Substituting:

$x-\overline{ x} = 40i$

Therefore,

Real and Complex Numbers

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