How to divide complex numbers - SAT Math

Card 0 of 8

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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