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SAT Math : SAT Mathematics

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Example Questions

Example Question #2381 : Sat Mathematics

Define an operation $\Theta$ such that for any complex number ,

$\Theta a = 2a i + i$

If $\Theta N = 1,000$ , evaluate .

Possible Answers:

$500 + \frac{1}{2}i$

$\frac{ 1 }{2}+ 500i$

$- \frac{ 1 }{2}- 500i$

$- \frac{ 1 }{2}+ 500i$

$500 - \frac{1}{2}i$

Correct answer:

$- \frac{ 1 }{2}- 500i$

Explanation:

First substitute our variable N in where ever there is an a.

Thus, $\Theta a = 2a i + i$ , becomes $\Theta N = 2N i + i$ .

Since $\Theta N = 1,000$ , substitute:

2N i + i = 1,000

In order to solve for the variable we will need to isolate the variable on one side with all other constants on the other side. To do this, apply the oppisite operation to the function.

First subtract i from both sides.

2N i= 1,000 - i

Now divide by 2i on both sides.

$\frac{2N i}{2i}= \frac{1,000 - i}{2i}$

From here multiply the numerator and denominator by i to further solve.

$N= \frac{1,000 - i}{2i}$

$= \frac{(1,000 - i) \cdot i }{2i \cdot i}$

$= \frac{1,000\cdot i - i\cdot i }{2 \cdot i^{2}}$

Recall that i^2=-1 by definition. Therefore,

$= \frac{1,000 i - (-1) }{2 \cdot (-1)}$

$= \frac{1,000 i +1 }{-2}$

$= \frac{ 1 }{-2}+ \frac{1,000 i }{-2}$

$=- \frac{ 1 }{2}- 500i$ .

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Example Question #1 : How To Divide Complex Numbers

Let 4x^2 = y^3 . What is the following equivalent to, in terms of :

$\frac{x}{\sqrt{y}}$

Possible Answers:

$\frac{1}{4}y^2$

$\pm\frac{1}{8}y^{\frac{3}{2}}$

$\frac{1}{2}\sqrt{y}$

$\pm\frac{1}{2}y$

Correct answer:

$\pm\frac{1}{2}y$

Explanation:

Solve for x first in terms of y, and plug back into the equation.

$\\4x^2 = y^3 \\ \\x^2 = \frac{1}{4} y^3 \\ \\x = \pm \frac{1}{2} y\sqrt{y}$

Then go back to the equation you are solving for:

$\frac{x}{\sqrt{y}}$ substitute in $x=\pm\frac{1}{2}y\sqrt{y}$

$\\ \frac{x}{\sqrt{y}}=\frac{\pm\frac{1}{2}y\sqrt{y}}{\sqrt{y}} \\ \\\frac{x}{\sqrt{y}}=\pm\frac{1}{2}y$

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Example Question #2 : How To Divide Complex Numbers

Simplify the expression by rationalizing the denominator, and write the result in standard form: $\frac{5+i}{7-i}$

Possible Answers:

$\frac{3}{4} + \frac{ 1}{4}i$

$\frac{18}{25} + \frac{ 6}{25}i$

$\frac{17}{24} + \frac{ 1}{4}i$

$\frac{17}{25} + \frac{ 6}{25}i$

$\left (\textup{Note: } i=\sqrt{-1}\right\ )$

Correct answer:

$\frac{17}{25} + \frac{ 6}{25}i$

Explanation:

Multiply both numerator and denominator by the complex conjugate of the denominator, which is 7 + i :

$\frac{5+i}{7-i}$

$= \frac{(5+i)(7+i)}{(7-i)(7+i)}$

$= \frac{5 \cdot 7 + 5 \cdot i + 7 \cdot i + i \cdot i }{(7^{2}+1 ^{2}) }$

$= \frac{35 +5i + 7 i +(-1) }{(49+1) }$

$= \frac{34 +12i }{50 }$

$= \frac{34 }{50 } + \frac{ 12i }{50 }$

$= \frac{17}{25} + \frac{ 6}{25}i$

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Example Question #2381 : Sat Mathematics

Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one.

Possible Answers:

24 + 7i

12 - 12i

7 + i

Correct answer: 24 + 7i

Explanation:

Distribute (3 + 4i)(4 - 3i)

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

12 - 9i + 16i -12i²

12 + 7i - 12(-1)

12 + 7i + 12

24 + 7i

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Example Question #2382 : Sat Mathematics

$x^{4}-1$ has 4 roots, including the complex numbers. Take the product of $\left ( 5+2i \right )$ with each of these roots. Take the sum of these 4 results. Which of the following is equal to this sum?

Possible Answers:

3+7i

4-10i

10i

The correct answer is not listed.

Correct answer:

Explanation:

$x^{4}-1=\left ( x^{2} -1\right )\left ( x^{2}+1 \right )=\left ( x+1 \right )\left ( x-1 \right )\left ( x-i \right )\left ( x+i \right )$

This gives us roots of

$1,\ -1,\ i,\ -i$

The product of $\left ( 5+2i \right )$ with each of these gives us:

$(5+2i)\cdot i = 5i + 2i^2 = 5i +2(-1) = 5i-2 = -2 + 5i$

$(5+2i)\cdot (-i) = -5i -2i^2 = -5i -2(-1) = -5i +2 = 2-5i$

$(5+2i)\cdot 1 = 5+2i$

$(5+2i)\cdot (-1) = -5 -2i$

The sum of these 4 is:

[(-2+5i) + (2-5i)] + [(5+2i) + (-5-2i)] =

=[(-2)+2 +5i - 5i] + [5-5 + 2i - 2i] = [0] + [0] = 0

What we notice is that each of the roots has a negative. It thus makes sense that they will all cancel out. Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property:

$1(5+2i) + (-1)\cdot(5+2i)+ i(5+2i)+(-i) \cdot(5+2i) = (5+2i)\cdot[1+(-1)+i+(-i)]$

$= (5+2i)\cdot[0] = 0$

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Example Question #2386 : Sat Mathematics

Simplify:

$(3i)^{6}$

Possible Answers:

729 i

-729 i

None of the other responses gives the correct answer.

729

Correct answer:

Explanation:

Apply the Power of a Product Property:

$(3i)^{6} = 3 ^{6} \cdot i^{6} = 729 \cdot i^{6}$

A power of can be found by dividing the exponent by 4 and noting the remainder. 6 divided by 4 is equal to 1, with remainder 2, so

$i^{6} = i^{2} = -1$

Substituting,

$(3i)^{6} = 729 \cdot (-1) = -729$ .

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Example Question #3 : How To Multiply Complex Numbers

Multiply $3 + i \sqrt {3 }$ by its complex conjugate.

Possible Answers:

None of the other responses gives the correct answer.

Correct answer:

Explanation:

The complex conjugate of a complex number a+ bi is a - bi . The product of the two is the number

$a^{2}+b^{2}$ .

Therefore, the product of $3 + i \sqrt {3 }$ and its complex conjugate $3 - i \sqrt {3 }$ can be found by setting a = 3 and $b = \sqrt{3}$ in this pattern:

$a^{2}+b^{2} = 3^{2}+ (\sqrt{3})^{2} = 9 + 3 = 12$ ,

the correct response.

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Example Question #2 : How To Multiply Complex Numbers

Multiply $11 - i \sqrt{11}$ by its complex conjugate.

Possible Answers:

110i

132 i

110

132

Correct answer:

132

Explanation:

The complex conjugate of a complex number a - bi is a+ bi . The product of the two is the number

$a^{2}+b^{2}$ .

Therefore, the product of $11 - i \sqrt{11}$ and its complex conjugate $11 + i \sqrt{11}$ can be found by setting a = 11 and $b = \sqrt{11}$ in this pattern:

$a^{2}+b^{2} = 11^{2}+ (\sqrt{11})^{2} = 121+11 = 132$ ,

the correct response.

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Example Question #2 : How To Multiply Complex Numbers

What is the product of $\frac{3}{2}+ \frac{1}{4} i$ and its complex conjugate?

Possible Answers:

$\frac{35}{16} i$

$\frac{35}{16}$

$-\frac{35}{16} i$

$-\frac{35}{16}$

The correct response is not among the other choices.

Correct answer:

The correct response is not among the other choices.

Explanation:

The complex conjugate of a complex number a+bi is a - bi , so $\frac{3}{2} + \frac{1}{4} i$ has $\frac{3}{2} - \frac{1}{4} i$ as its complex conjugate.

The product of a+bi and a - bi is equal to $a^{2}+ b^{2}$ , so set $a = \frac{3}{2} , b= \frac{1}{4}$ in this expression, and evaluate:

$a^{2}+ b^{2} = \left ( \frac{3}{2} \right )^{2} + \left (\frac{1}{4} \right )^{2} = \frac{9}{4} + \frac{1}{16} = \frac{37}{16}$ .

This is not among the given responses.

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Example Question #1 : How To Multiply Complex Numbers

Multiply and simplify:

$\sqrt{-35} \cdot \sqrt{-5}$

Possible Answers:

$5i \sqrt{7}$

$7i \sqrt{5}$

$-7i \sqrt{5}$

$-5i \sqrt{7}$

None of the other choices gives the correct response.

Correct answer:

None of the other choices gives the correct response.

Explanation:

The two factors are both square roots of negative numbers, and are therefore imaginary. Write both in terms of before multiplying:

$\sqrt{-35} = i \sqrt{35}$

$\sqrt{-5} = i \sqrt{5}$

Therefore, using the Product of Radicals rule:

$\sqrt{-35} \cdot \sqrt{-5}$

$= i \sqrt{35} \cdot i \sqrt{5}$

$= i \cdot i \cdot \sqrt{35} \cdot \sqrt{5}$

$= i ^{2}\cdot \sqrt{35 \cdot 5 }$

$=-1 \cdot \sqrt{175 }$

$=-1 \cdot \sqrt{25 }\cdot \sqrt{7}$

$=-1 \cdot 5\cdot \sqrt{7}$

$=-5\sqrt{7}$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2381 : Sat Mathematics

Example Question #1 : How To Divide Complex Numbers

Example Question #2 : How To Divide Complex Numbers

Example Question #2381 : Sat Mathematics

Example Question #2382 : Sat Mathematics

Example Question #2386 : Sat Mathematics

Example Question #3 : How To Multiply Complex Numbers

Example Question #2 : How To Multiply Complex Numbers

Example Question #2 : How To Multiply Complex Numbers

Example Question #1 : How To Multiply Complex Numbers

All SAT Math Resources

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