Complex Numbers - SAT Math

Card 0 of 344

Question

Let . What is the following equivalent to, in terms of :

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

Answer

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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Question

For which of the following values of is the value of ${x^{-\frac{1}{2}}}$ least?

Answer

${x^{-\frac{1}{2}}}$ is the same as $\frac{1}{\sqrt{x}}$ , which means that the bigger the answer to $\sqrt{x}$ is, the smaller the fraction will be.

Therefore, is the correct answer because

$\frac{1}{\sqrt{16}} = \frac{1}{4}$ .

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Question

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(2+i) \ast i$ .

Answer

$a \ast b = \frac{a + b}{a}$ , so

$(2+i) \ast i = \frac{(2+ i )+ i}{2+i} = \frac{2 + 2i}{2+i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$(2+i) \ast i$

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

$= \frac{\left (2 + 2i \right )\left (2-i \right )}{\left (2+i \right )\left (2-i \right )}$

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

$= \frac{ 4 - 2 i + 4i - 2 \cdot i^{2}}{4+1}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot (-1)}{5}$

$= \frac{ 4 - 2 i + 4i +2}{5}$

$= \frac{ 6 + 2 i }{5}$

$= \frac{ 6 }{5} + \frac{ 2 }{5}i$

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Question

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

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

Answer

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

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

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

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

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

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

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

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

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Question

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(3+2i) \ast 2$

Answer

$a \ast b = \frac{a + b}{a}$ , so

$(3+2i) \ast 2 = \frac{(3+2i) + 2}{3+2i } = \frac{5+2i}{3+2i }$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$(3+2i) \ast 2 = \frac{5+2i}{3+2i }$

$= \frac{\left (5+2i \right )\left (3-2i \right )}{\left (3+2i \right )\left (3-2i \right )}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2i \cdot 2i }{3^{2}+2^{2}}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2 \cdot 2 \cdot i^{2} }{9+4}$

$= \frac{15 - 10i + 6i -4 \cdot (-1) }{9+4}$

$= \frac{15 - 10i + 6i +4}{13}$

$= \frac{19 -4i}{13}$

$= \frac{19 }{13} - \frac{ 4}{13}i$

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Question

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

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

Answer

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

, and

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

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

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

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

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

$=\frac{ i - 2i^{2}}{1+4}$

$=\frac{ i - 2(-1)}{5}$

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

$=\frac{ 2}{5} + \frac{1}{5} i$

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Question

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

$\Theta a = 2a i + i$

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

Answer

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:

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.

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 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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Question

Define an operation $\blacklozenge$ as follows:

For any two complex numbers and ,

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$

Evaluate $(1+ i) \blacklozenge (1+ i)$ .

Answer

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$ , so

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

We can simplify each expression separately by rationalizing the denominators.

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

$= \frac{\left (1+ i \right )\cdot i}{2i \cdot i}$

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

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

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

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

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

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

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

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

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

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

Therefore,

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

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

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

$= \frac{ 3 }{2} + \frac{1}{2} i$

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Question

From $\frac{3}{2} + \frac{1}{4} i$ , subtract its complex conjugate. What is the difference ?

Answer

The complex conjugate of a complex number is , so $\frac{3}{2} + \frac{1}{4} i$ has $\frac{3}{2} - \frac{1}{4} i$ as its complex conjugate. Subtract the latter from the former:

$\left (\frac{3}{2} + \frac{1}{4} i \right ) - \left ( \frac{3}{2} - \frac{1}{4} i \right )$

$= \frac{3}{2} + \frac{1}{4} i - \frac{3}{2}+ \frac{1}{4} i$

$= \frac{3}{2}- \frac{3}{2} + \frac{1}{4} i + \frac{1}{4} i$

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

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Question

From $17 - i \sqrt{17}$ , subtract its complex conjugate.

Answer

The complex conjugate of a complex number is . Therefore, the complex conjugate of $17 - i \sqrt{17}$ is $17 + i \sqrt{17}$ ; subtract the latter from the former by subtracting real parts and subtracting imaginary parts, as follows:

$(17 - i \sqrt{17})- (17 + i \sqrt{17})$

$= 17 - i \sqrt{17} - 17 - i \sqrt{17}$

$= 17 - 17 - i \sqrt{17} - i \sqrt{17}$

$= - 2 i \sqrt{17}$

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Question

From $7 + i \sqrt {7 }$ , subtract its complex conjugate.

Answer

The complex conjugate of a complex number is . Therefore, the complex conjugate of $7 + i \sqrt {7 }$ is $7 - i \sqrt {7 }$ ; subtract the latter from the former by subtracting real parts and subtracting imaginary parts, as follows:

$(7 + i \sqrt {7 })- (7 - i \sqrt {7 })$

$= 7- 7 + i \sqrt {7 }- (- i \sqrt {7 })$

$= i \sqrt {7 }+ i \sqrt {7 }$

$=2i \sqrt {7 }$

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Question

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

Answer

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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Question

$x^{4}-1$ 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?

Answer

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

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:

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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Question

Simplify:

$(3i)^{6}$

Answer

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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Question

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

Answer

The complex conjugate of a complex number is . 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 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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Question

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

Answer

The complex conjugate of a complex number is . 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 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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Question

Evaluate:

$i^{-31}$

Answer

$a^{-m}$ is defined to be equal to $\frac{1}{a^{m}}$ for any real or imaginary and for any real ; therefore,

$i^{-31} = \frac{1}{i^{31}}$

To evaluate a positive power of , divide the power by 4 and note the remainder:

$31 \div 4= 7 \textrm{ R }3$

Therefore,

$i ^{31} = i ^{3} = -i$

Substituting,

$i^{-31} = \frac{1}{-i}$

Rationalizing the denominator by multiplying both numerator and denominator by :

$i^{-31} = \frac{1 \cdot i}{-i \cdot i }= \frac{ i}{-i ^{2}}= \frac{ i}{-(-1)}= \frac{ i}{1}= i$

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Question

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

Answer

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

The product of and 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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Question

Multiply and simplify:

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

Answer

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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Question

Evaluate $x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}$

Answer

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}$ is recognizable as the cube of the binomial . That is,

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}= (x-y) ^{3}$

Therefore, setting and and evaluating:

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}=[ (7+4i) - (7-4 i) ]^{3}$

$= (7-7 +4i+4i ) ^{3}$

$=(8i)^{3}$

Applying the Power of a Product Rule and the fact that $i^{3} = -i$ :

$(8i)^{3} = 8 ^{3} i ^{3} = 512 (-i) = -512i$ ,

the correct value.

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