Real and Complex Numbers

SAT Subject Test in Math I · Learn by Concept

Help Questions

SAT Subject Test in Math I › Real and Complex Numbers

1 - 10

Which answer choice has the greatest real number value?

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

CORRECT

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

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

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

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

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.

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

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

Evaluate .

CORRECT

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,

Evaulate:

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

CORRECT

Explanation

Multiply both numerator and denominator by , then divide termwise:

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

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

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

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

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

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

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

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

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

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

None of these

CORRECT

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,

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

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

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

which is not among the choices.

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

All of these

CORRECT

$1 - i \sqrt{3 }$

$-1 + i \sqrt{3 }$

$-1 - i \sqrt{3 }$

$1 + i \sqrt{3 }$

Explanation

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.

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

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

CORRECT

None of these

Explanation

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,

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

What is the conjugate for the complex number

CORRECT

Explanation

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.