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ACT Math : How to divide complex numbers

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Example Questions

Example Question #1 : How To Divide Complex Numbers

Simplify: $\frac{9+4i}{4i}$

Possible Answers:

$\frac{ 9 }{4 } - i$

$\frac{ 9 }{4 } + i$

$1+ \frac{ 9 }{4 } i$

$1- \frac{ 9 }{4 } i$

$-\frac{ 9 }{4 } + i$

Correct answer:

$1- \frac{ 9 }{4 } i$

Explanation:

Multiply both numberator and denominator by :

$\frac{9+4i}{4i}$

$= \frac{\left (9+4i \right )\cdot i}{4i \cdot i}$

$= \frac{9i + 4i^{2}}{4 i^{2}}$

$= \frac{9i + 4 (-1)}{4 (-1)}$

$= \frac{-4+9i }{-4 }$

$= \frac{-4 }{-4 }+ \frac{ 9i }{-4 }$

$= 1- \frac{ 9 }{4 } i$

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Example Question #31 : Squaring / Square Roots / Radicals

Evaluate: $100 \div 5i \div 5i$

Possible Answers:

-4i

100i

Correct answer:

Explanation:

First, divide 100 by as follows:

$\frac{100}{5i} = \frac{100 \cdot i}{5i \cdot i} = \frac{100 i}{5(-1)} = \frac{100 i}{-5}= -20i$

Now dvide this result by :

$\frac{-20i}{5i} = \frac{-20}{5} = -4$

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Example Question #3 : How To Divide Complex Numbers

Evaluate: $100 \div (1+i) \div (1-i)$

Possible Answers:

-25 +25i

25+ 25i

25-25i

Correct answer:

Explanation:

First, divide 100 by 1+i as follows:

$\frac{100}{1+i}$

$=\frac{100(1-i)}{\left (1+i \right )(1-i)}$

$= \frac{100-100i}{1^{2}+1^{2}}$

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

= 50-50i

Now, divide this by 1-i :

$\frac{ 50-50i}{1-i} = \frac{ 50 (1-i)}{1-i} = 50$

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Example Question #4 : How To Divide Complex Numbers

Evaluate: $100i \div (4i )^{2}$

Possible Answers:

$\frac{25}{4}i$

$-\frac{25}{2}$

$\frac{25}{2}i$

$- \frac{25}{4}i$

$- \frac{25}{4}$

Correct answer:

$- \frac{25}{4}i$

Explanation:

First, evaluate $(4i)^{2}$ :

$\left (4i \right )^{2} = 16 i^{2} = 16 (-1)= -16$

Now divide this into 100 i :

$\frac{100 i}{-16} =- \frac{25}{4}i$

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Example Question #5 : How To Divide Complex Numbers

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

Possible Answers:

$\frac{1}{2} i$

$\frac{1}{2}$

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

Correct answer:

$\frac{1}{2}$

Explanation:

First, evaluate $\left ( 1+ i\right )^{2}$ using the square pattern:

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

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

=1+2i+ (-1)

=2i

Divide this into :

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

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Example Question #1 : How To Divide Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify: $\frac{6+12i}{4}$

Possible Answers:

$\frac{3}{2} + \frac{12i}{4i}$

6 + 3i

$\textup{None of these answers are correct}$

$\frac{3}{2} + 12i$

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

Correct answer:

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

Explanation:

This problem can be solved very similarly to a binomial such as a+bx . In this case, both the real and nonreal terms in the complex number are eligible to be divided by the real divisor.

$\frac{6+12i}{4} = (\frac{6}{4} + \frac{12i}{4})$

$\frac{12i}{4} = \frac{12}{4} \cdot i = 3 \cdot i = 3i$ , so

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

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Example Question #7 : How To Divide Complex Numbers

Complex numbers take the form a+bi , where is the real term in the complex number and is the nonreal (imaginary) term in the complex number.

Simplify by using conjugates: $\frac{4+2i}{-3-2i}$

Possible Answers:

$\frac{-16-2i}{13}$

$\frac{-16+2i}{-13}$

$\frac{-8-2i}{5}$

$\frac{-8-2i}{13}$

$\frac{-16-2i}{5}$

Correct answer:

$\frac{-16-2i}{13}$

Explanation:

Solving this problem using a conjugate is just like conjugating a binomial to simplify a denominator.

$\frac{(4+2i)(-3+2i)}{(-3-2i)(-3+2i)}$ Multiply both terms by the denominator's conjugate.

$\frac{(4+2i)(-3+2i)}{(-3-2i)(-3+2i)} = \frac{(4+2i)(-3+2i)}{13}$ Simplify. Note .

Combine and simplify.

Simplify the numerator.

$\frac{-16-2i}{13}$ The prime denominator prevents further simplifying.

Thus, $\frac{4+2i}{-3-2i} = \frac{-16-2i}{13}$ .

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Example Question #8 : How To Divide Complex Numbers

Simplify:

${}\frac{5+7i}{2+i}$

Possible Answers:

$\frac{5}{7}+2i{}$

$\frac{5}{2}+\frac{7}{2}i{}$

$\frac{17}{5}+\frac{9}{5}i{}$

1+3i{}

Correct answer:

$\frac{17}{5}+\frac{9}{5}i{}$

Explanation:

This problem can be solved in a way similar to other kinds of division problems (with binomials, for example). We need to get the imaginary number out of the denominator, so we will multiply the denominator by its conjugate and multiply the top by it as well to preserve the number's value.

$\frac{5+7i}{2+i} \cdot \frac{2-i}{2-i}=\frac{10-5i+14i-7i^2}{4-i^2}{}$

Then, recall i^2 = -1 by definition, so we can simplify this further:

$\frac{10+7+14i-5i}{4+1} = \frac{17}{5}+\frac{9}{5}i{}$

This is as far as we can simplify, so it is our final answer.

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ACT Math : How to divide complex numbers

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Divide Complex Numbers

Example Question #31 : Squaring / Square Roots / Radicals

Example Question #3 : How To Divide Complex Numbers

Example Question #4 : How To Divide Complex Numbers

Example Question #5 : How To Divide Complex Numbers

Example Question #1 : How To Divide Complex Numbers

Example Question #7 : How To Divide Complex Numbers

Example Question #8 : How To Divide Complex Numbers

All ACT Math Resources

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