How to divide complex numbers - PSAT Math

Card 0 of 63

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

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

Simplify by rationalizing the denominator:

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

Answer

Multiply both numerator and denominator by :

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

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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