Complex Numbers - PSAT Math

Card 0 of 20

Question

Simplify by rationalizing the denominator:

$\frac{20 + 6i}{8i}$

Answer

Multiply both numerator and denominator by :

$\frac{20 + 6i}{8i}$

$=\frac{\left (20 + 6i \right ) \cdot i}{8i \cdot i}$

$=\frac{ 20i + 6i^{2} }{8i ^{2}}$

$=\frac{ 20i + 6(-1) }{8 (-1)}$

$=\frac{ 20i -6 }{-8}$

$=\frac{ -6+ 20i }{-8}$

$=\frac{ -6 }{-8} + \frac{ 20i }{-8}$

$=\frac{ 3 }{4} - \frac{ 5}{2}i$

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Question

Divide:

$-100 \div 5i$

Answer

Rewrite in fraction form, and multiply both numerator and denominator by :

$-100 \div 5i$

$= \frac{-100}{5i}$

$= \frac{-100 \cdot i }{5i \cdot i }$

$= \frac{-100 i }{5i ^{2}}$

$= \frac{-100 i }{5 (-1)}$

$= \frac{-100 i }{-5}$

$= \frac{-100 }{-5}\cdot i$

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Question

Simplify by rationalizing the denominator:

$\frac{ 25- 12i}{ - 10i}$

Answer

Multiply both numerator and denominator by :

$\frac{ 25- 12i}{ - 10i}$

$=\frac{\left ( 25- 12i \right )\cdot i}{ - 10i \cdot i }$

$=\frac{ 25i- 12i ^{2} }{ - 10i ^{2} }$

$=\frac{ 25i- 12 (-1)}{ - 10 (-1) }$

$=\frac{ 25i+ 12}{ 10}$

$=\frac{12+ 25i }{ 10}$

$=\frac{12 }{ 10} + \frac{ 25i }{ 10}$

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

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Question

Simplify by rationalizing the denominator:

$\frac{ 21i}{3+ i\sqrt{5}}$

Answer

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is $3- i\sqrt{5}$ :

$\frac{ 21i}{3+ i\sqrt{5}}$

$= \frac{ 21i\left (3- i\sqrt{5} \right )}{\left (3+ i\sqrt{5} \right )\left (3- i\sqrt{5} \right )}$

$= \frac{21i \cdot 3- 21i \cdot i\sqrt{5} }{ 3^{2}+ \left ( \sqrt{5} \right )^{2}}$

$= \frac{ 63i - 21i^{2} \sqrt{5}}{ 9+5}$

$= \frac{ 63i - 21 (-1)\sqrt{5}}{ 14}$

$= \frac{ 21 \sqrt{5}+ 63i }{ 14}$

$= \frac{ 21 \sqrt{5} }{ 14}+ \frac{63 i}{ 14}$

$= \frac{ 3\sqrt{5} }{ 2} + \frac{ 9 }{ 2} i$

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Question

Simplify by rationalizing the denominator:

$\frac{2+3i}{4+ 3i}$

Answer

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

$\frac{2+3i}{4+ 3i}$

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

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

$=\frac{8 - 6i +12i - 9 i ^{2}}{16+ 9}$

$=\frac{8 - 6i +12i - 9(-1)}{25}$

$=\frac{8 - 6i +12i +9}{25}$

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

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

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Question

Simplify by rationalizing the denominator:

$\frac{15i}{4- 3i}$

Answer

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

$\frac{15i}{4- 3i}$

$= \frac{15i\left (4+ 3i \right )}{\left (4- 3i \right )\left (4+ 3i \right )}$

$= \frac{15i\cdot 4+ 15i\cdot 3i }{4^{2}+3^{2}}$

$= \frac{60 i + 45i^{2} }{16+9}$

$= \frac{60 i + 45 (-1)}{25}$

$= \frac{-45 + 60 i}{25}$

$= \frac{-45 }{25} + \frac{ 60 i}{25}$

$= - \frac{9}{5} + \frac{ 12}{5}i$

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Question

Define an operation $\ast$ as follows:

For all complex numbers ,

$a \ast b = \frac{ab}{i}$ .

Evaluate $(3+ 4i) \ast (4-3i)$ .

Answer

$a \ast b = \frac{ab}{i}$

$(3+ 4i) \ast (4-3i)= \frac{(3+ 4i) (4-3i)}{i}$

$= \frac{ 3 \cdot 4 - 3 \cdot 3i + 4i \cdot 4 -4i \cdot 3i}{i}$

$= \frac{ 12 - 9i + 16i -12i ^{2}}{i}$

$= \frac{ 12 - 9i + 16i -12 (-1)}{i}$

$= \frac{ 12 - 9i + 16i +12}{i}$

$= \frac{ 24+7i }{i}$

$= \frac{ \left (24+7i \right ) (-i)}{i (-i)}$

$= \frac{ -24i-7i ^{2} }{ -i^{2}}$

$= \frac{ -24i-7(-1) }{ -(-1)}}$

$= \frac{ -24i+7 }{ 1}$

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Question

Define an operation $\ast$ as follows:

For all complex numbers ,

$a \ast b = \frac{ai}{b}$ .

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

Answer

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

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

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

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

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

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is . The above becomes:

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

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

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

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

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

$= \frac{-4 +3i }{5}$

$=- \frac{4}{5} +\frac {3 }{5}i$

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Question

Define an operation $\ast$ as follows:

For all complex numbers ,

$a \ast b = \frac{ a^{3}}{b^{3} + 1}$

Evaluate $i \ast i$ .

Answer

$a \ast b = \frac{ a^{3}}{b^{3} + 1}$

$i \ast i = \frac{ i^{3}}{i^{3} + 1}$

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

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

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

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is . The above becomes:

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

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

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

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

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

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

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

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Question

Which of the following is equal to $i ^{-56}$ ?

Answer

$i ^{-56} = \frac{1}{i^{56}}$

$56 \div 4 = 14$ , so 56 is a multiple of 4. raised to the power of any multiple of 4 is equal to 1, so

$i ^{-56} = \frac{1}{i^{56}} = \frac{1}{1} = 1$ .

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Question

Multiply:

Answer

This is the product of a complex number and its complex conjugate. They can be multiplied using the pattern

$\left (A + Bi \right )\left (A - Bi \right ) = A^{2} + B^{2}$

with

$(10 + 3i)(10-3i) = 10^{2} +3^{2} = 100 + 9 = 109$

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Question

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

Answer

By the power of a product property,

$\left (3i \right )^{5} = 3^{5} \cdot i^{5} = 3^{5} \cdot i^{4} \cdot i = 243 \cdot 1 \cdot i = 243 i$

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Question

Which of the following is equal to ?

Answer

The first step to solving this problem is distributing the exponent:

Next, we need simplify the complex portion.

$i^7=i^2\times i^2\times i^2\times i=-1\times -1\times -1\times i=-i$

Thus, our final answer is $128\times (-i)=-128i$ .

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Question

Simplify:

$\small \left ( 3 - 2i \right )\left ( 2 + i\right )$

Answer

Use the FOIL method that states to multiply the Firsts, Outter, Inner, Lasts. Also remember that $\small i \times i = -1$ :

$\small \left ( 3 - 2i \right )\left ( 2 + i\right )$

$\small = 3 \times 2 + 3 \times i - 2i \times 2 - 2i \times i$

$\small = 6 +3i -4i + 2$

$\small = 8 - i$

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Question

What is the eighth power of ?

Answer

$\left (3i \right )^{8}$

$= 3^{8} i^{8}$

$= 6,561 i^{8}$

raised to the power of any multiple of 4 is equal to 1, so the above expresion is equal to

$= 6,561 \cdot 1 = 6,561$

This is not among the given choices.

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Question

What is the third power of $2 + i\sqrt{5}$ ?

Answer

You are being asked to evaluate

$\left (2 + i\sqrt{5} \right )^{3}$

You can use the cube of a binomial pattern with $A = 2, B = i\sqrt{5}$ :

$\left ( A + B \right )^{3} = A^{3}+ 3A^{2}B + 3AB^{2}+ B^{3}$

$\left ( 2 + i\sqrt{5} \right )^{3} =2^{3}+ 3 \cdot 2^{2} \cdot i\sqrt{5} + 3 \cdot 2 (i\sqrt{5}) ^{2}+ (i\sqrt{5}) ^{3}$

$=8+ 3 \cdot 4 \cdot i\sqrt{5} + 6i ^{2} (\sqrt{5}) ^{2}+ i^{3}(\sqrt{5}) ^{3}$

$=8+12i\sqrt{5} + 6 (-1) 5+(-i)(5\sqrt{5})$

$=8-30 +12i\sqrt{5} -5i\sqrt{5}$

$=-22 +7i\sqrt{5}$

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Question

What is the fourth power of $3 - i\sqrt{2}$ ?

Answer

$\left (3 - i\sqrt{2} \right )^{4}$ can be calculated by squaring $3 - i\sqrt{2}$ , then squaring the result, using the square of a binomial pattern as follows:

$\left (3 - i\sqrt{2} \right )^{2}$

$=3^{2}- 2 \cdot 3 \cdot i\sqrt{2} +\left ( i\sqrt{2} \right )^{2}$

$=9- 6 i\sqrt{2} +2i^{2}$

$=9- 6 i\sqrt{2} +2(-1)$

$=7- 6 i\sqrt{2}$

$\left (3 - i\sqrt{2} \right )^{4}$

$=\left [ \left (3 - i\sqrt{2} \right )^{2} \right ]^{2}$

$=\left (7- 6 i\sqrt{2}} \right )^{2}$

$=7^{2} - 2 \cdot 7 \cdot 6 i\sqrt{2}} + \left ( 6 i\sqrt{2}} \right )^{2}$

$=49-84 i\sqrt{2}} + 6^{2} i^{2} \left ( \sqrt{2}} \right )^{2}$

$=49-84 i\sqrt{2}} +36 (-1) (2)$

$=49-84 i\sqrt{2}} -72$

$=-23-84 i\sqrt{2}}$

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Question

Multiply by its complex conjugate. What is the product?

Answer

The product of any complex number and its complex conjugate is the real number $a^{2}+b^{2}$ , so all that is needed here is to evaluate the expression:

$32^{2}+ 18 ^{2}= 1,024+ 324 = 1,348$

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Question

What is the eighth power of ?

Answer

First, square using the square of a binomial pattern as follows:

$\left (1+ i \right )^{2}$

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

Raising this number to the fourth power yields the correct response:

$\left (1+ i \right )^{8}$

$= \left [\left (1+ i \right )^{2} \right ]^{4}$

$= (2i)^{4}$

$= (2 )^{4} \cdot i^{4}$

$=16 \cdot 1 = 16$

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Question

What is the ninth power of ?

Answer

$\left ( -2i\right )^{9}$

$= \left ( -2 \right )^{9} \cdot i^{9}$

To raise a negative number to an odd power, take the absolute value of the base to that power and give its opposite:

$\left ( -2\right )^{9} = - \left ( 2^{9}\right ) = -512$

To raise to a power, divide the power by 4 and raise to the remainder. Since

$9 \div 4 = 2 \textup{ R }1$ ,

$i^{9} = i^{1}= i$

Therefore,

$\left ( -2i\right )^{9}= \left ( -2 \right )^{9} \cdot i^{9} = -512 i$

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