Real and Complex Numbers - SAT Subject Test in Math I

Card 0 of 13

Question

Evaulate:

$\frac{45-35i}{5i}$

Answer

Multiply both numerator and denominator by , then divide termwise:

$\frac{45-35i}{5i}$

$= \frac{\left (45-35i \right )\cdot i}{5i \cdot i}$

$= \frac{ 45\cdot i-35i\cdot i }{5i ^{2}}$

$= \frac{ 45 i-35i ^{2}}{5i ^{2}}$

$= \frac{ 45 i-35(-1) }{5 (-1)}$

$= \frac{ 45 i- (-35) }{-5 }$

$= \frac{35+ 45 i }{-5 }$

$= \frac{35 }{-5 } + \frac{ 45 i }{-5 }$

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Question

Which of the following is equal to $\left (5i \right ) ^{3}$ ?

Answer

By the power of a product property,

$\left (5i \right ) ^{3} = 5^{3} \cdot i^{3} = 125(-i)= -125i$

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Question

Multiply:

$3 i \cdot 5i \cdot 7i \cdot 2i$

Answer

$3 i \cdot 5i \cdot 7i \cdot 2i$

$= 3 \cdot 5 \cdot 7 \cdot 2 \cdot i \cdot i \cdot i \cdot i$

$= 210 \cdot i ^{4} = 210 \cdot 1 = 210$

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Question

Which of the following is equal to $i^{-33}$ ?

Answer

$i^{-33} = \frac{1}{i^{33}}$

To raise to a power, divide the exponent by 4, note its remainder, and raise to the power of that remainder:

$33 \div 4 = 8 \textrm{ R }1$

$i^{33} = i ^{1} = i$

Therefore, $i^{-33} = \frac{1}{i^{33}} = \frac{1}{i} = \frac{1 \cdot i}{i\cdot i} = \frac{i}{-1} = -i$

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Question

Which of the following is equal to $\left (2i \right )^{6}$ ?

Answer

By the power of a product property,

$\left (2i \right )^{6} = 2^{6} \cdot i^{6} = 2^{6} \cdot i^{4} \cdot i^{2} = 64 \cdot 1 \cdot (-1) = -64$

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Question

What is the conjugate for the complex number

Answer

To find the conjugate of the complex number of the form $\small a\pm bi$ , change the sign on the complex term. The complex part of the problem is $\small -2i$ so changing the sign would make it a $\small +2i$ . The sign in the real part of the number, the 3 in this case, does not change sign.

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Question

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

If , then evaluate $x^{3} \cdot \overline{x} ^{3}$ .

Answer

Applying the Power of a Product Rule:

$x^{3} \cdot \overline{x} ^{3}=(x \overline{x})^{3}$

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

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

, so, setting in the above pattern:

$x \overline{x} = 5^{2} + 2^{2} = 25 + 4 = 29$

Consequently,

$x^{3} \cdot \overline{x} ^{3}=(x \overline{x})^{3} = 29 ^{3}= 24, 389$

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Question

Which of the following choices gives a sixth root of sixty-four?

Answer

Let be a sixth root of 64. The question is to find a solution of the equation

$x^{6} = 64$ .

Subtracting 64 from both sides, this equation becomes

$x^{6} - 64= 0$

64 is a perfect square (of 8) The binomial at left can be factored first as the difference of two squares:

$(x^{3} ) ^{2}- 8 ^{2}= 0$

$( x^{3} + 8 )( x^{3} - 8 )= 0$

8 is a perfect cube (of 2), so the two binomials can be factored as the sum and difference, respectively, of two cubes:

$x^{3} + 8 = x^{3}+ 2 ^{3} = (x+2) (x^{2}- x \cdot 2 + 2 ^{2}) = (x+2) (x^{2}-2x+4)$

$x^{3} - 8 = x^{3}+ 2 ^{3} = (x-2) (x^{2}+x \cdot 2 + 2 ^{2}) = (x-2) (x^{2}+2x+4)$

The equation therefore becomes

$(x+2) (x^{2}-2x+4) (x-2) (x^{2}+2x+4) = 0$ .

By the Zero Product Principle, one of these factors must be equal to 0.

If , then ; if , then . Therefore, and 2 are sixth roots of 64. However, these are not choices, so we examine the other polynomials for their zeroes.

If $x^{2}-2x+4 = 0$ , then, setting in the following quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$= \frac{- (-2) \pm \sqrt{ (-2)^{2}-4(1)(4)}}{2(1)}$

$= \frac{2 \pm \sqrt{ 4 - 16 }}{2}$

$= \frac{2 \pm \sqrt{ -12 }}{2}$

$= \frac{2 \pm\sqrt{4 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$

$= \frac{2 \pm 2i \sqrt{3 }}{2}$

$=1 \pm i \sqrt{3 }$

If $x^{2}+2x+4= 0$ , then, setting in the quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$= \frac{- 2 \pm \sqrt{ 2^{2}-4(1)(4)}}{2(1)}$

$= \frac{-2 \pm \sqrt{ 4 - 16 }}{2}$

$= \frac{-2 \pm \sqrt{ -12 }}{2}$

$= \frac{-2 \pm\sqrt{4 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$

$= \frac{-2 \pm 2i \sqrt{3 }}{2}$

$=-1 \pm i \sqrt{3 }$

Therefore, the set of sixth roots of 64 is

$\left { -4, 4,-1 - i \sqrt{3 }, -1 + i \sqrt{3 } ,1 - i \sqrt{3 }, 1 + i \sqrt{3 } \right }$ .

All four choices appear in this set.

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Question

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

If , then evaluate $x^{2} - \overline{x}^{2}$ .

Answer

By the difference of squares pattern,

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

If , then $\overline{x} = 5- 2i$ . As a result:

$x+\overline{ x} = (5+2i) + (5- 2i) = 5+ 5 + 2i - 2i = 10$

$x-\overline{ x} = (5+2i) - (5- 2i) = 5- 5 + 2i +2i = 4 i$

Therefore,

$x^{2} - \overline{x}^{2}$

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

$= 10 \cdot 4i$

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Question

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

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

Evaluate .

Answer

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,

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Question

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

Evaluate $\overline{x} \cdot \overline{y}$ .

Answer

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,

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

, so $\overline{xy }= 13- 7i$ , and

$\overline{x} \cdot \overline{y} =13- 7i$ ,

which is not among the choices.

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Question

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

.

Evaluate $\overline{x} - \overline{y}$ .

Answer

Let , where all variables represent real quantities.

Then

Since

,

if follows that

and

Also, by definition,

$\overline{x} = a- bi, \overline{y} = c-di$

$\overline{x} - \overline{y} = (a-bi) - (c- di)$

$\overline{x} - \overline{y} = a - c-bi + di$

$\overline{x} - \overline{y} =( a - c)+ (-b + d)i$

It is known that and , but without further information, nothing can be determined about 0r . Therefore, $\overline{x} - \overline{y}$ cannot be evaluated.

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Question

Which answer choice has the greatest real number value?

Answer

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.

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