How to divide complex numbers - PSAT Math

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Question

Simplify by rationalizing the denominator:

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

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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:

$-100 \div 5i$

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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:

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

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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}$}$

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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:

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

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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:

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

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

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

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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 $i \ast i$ .

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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-(-1) }{ 1+ 1 }$$

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

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

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

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Question

Simplify by rationalizing the denominator:

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

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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:

$-100 \div 5i$

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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:

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

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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}$}$

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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:

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

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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:

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

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

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

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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 $i \ast i$ .

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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-(-1) }{ 1+ 1 }$$

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

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

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

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Question

Simplify by rationalizing the denominator:

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

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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:

$-100 \div 5i$

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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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