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SAT Math : How to divide complex numbers

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Example Questions

例子问题# 21:Squaring / Square Roots / Radicals

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

Possible Answers:

$\frac{1}{4}$

$\frac{1}{16}$

Correct answer:

Explanation:

${x^{-\frac{1}{2}}}$ 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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Example Question #22 :Squaring / Square Roots / Radicals

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$(2+i) \ast i$

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

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

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

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

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

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

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

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

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Example Question #23 :Squaring / Square Roots / Radicals

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

$\Phi a = a + 2ai$

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

Possible Answers:

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

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

$\frac{ 2}{5} - \frac{1}{5} i$

$\frac{ 1}{5} - \frac{2}{5} i$

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

Correct answer:

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

Explanation:

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

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

$\Phi N = i$ , so

N (1+ 2i) = i , and

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

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 1 -2i :

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

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

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

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

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

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

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

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Example Question #24 :Squaring / Square Roots / Radicals

Define an operation $\blacklozenge$ as follows:

For any two complex numbersand,

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

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

Possible Answers:

$\frac{ 3 }{2} - \frac{1}{2} i$

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

$\frac{ 1 }{2} -\frac{3}{2} i$

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

Correct answer:

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

Explanation:

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

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

We can simplify each expression separately by rationalizing the denominators.

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

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

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

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

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

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

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

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

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

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

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

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

= 1 + i

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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Example Question #25 :Squaring / Square Roots / Radicals

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

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

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

Possible Answers:

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

$\frac{19}{13} - \frac{ 16}{13}i$

$\frac{11 }{13} - \frac{ 4}{13}i$

$\frac{11 }{13} + \frac{ 16}{13}i$

$\frac{19}{13} + \frac{ 16}{13}i$

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #26 :Squaring / Square Roots / Radicals

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

$\Theta a = 2a i + i$

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

Possible Answers:

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

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

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

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

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

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 #27 :Squaring / Square Roots / Radicals

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

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

Possible Answers:

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

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

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

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

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 #1 :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:

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

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

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

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

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

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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SAT Math : How to divide complex numbers

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

例子问题# 21:Squaring / Square Roots / Radicals

Example Question #22 :Squaring / Square Roots / Radicals

Example Question #23 :Squaring / Square Roots / Radicals

Example Question #24 :Squaring / Square Roots / Radicals

Example Question #25 :Squaring / Square Roots / Radicals

Example Question #26 :Squaring / Square Roots / Radicals

Example Question #27 :Squaring / Square Roots / Radicals

Example Question #1 :How To Divide Complex Numbers

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