How to divide complex numbers - ACT Math

Card 0 of 72

Question

Simplify:

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

Answer

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 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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

First, divide 100 by as follows:

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

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

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

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

Now, divide this by :

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

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Question

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

Answer

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

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

Now divide this into :

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

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Question

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

Answer

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

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

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

Divide this into :

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

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Question

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

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

Answer

This problem can be solved very similarly to a binomial such as . 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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Question

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

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

Answer

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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Question

Simplify:

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

Answer

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 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.

Compare your answer with the correct one above

Question

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

Answer

Multiply both numberator and denominator by :

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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 :

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

Compare your answer with the correct one above

Question

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

Answer

First, divide 100 by as follows:

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

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

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

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

Now, divide this by :

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Now divide this into :

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

Divide this into :

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

Compare your answer with the correct one above

Question

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

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

Answer

This problem can be solved very similarly to a binomial such as . 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

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

Compare your answer with the correct one above

Question

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

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

Answer

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

Compare your answer with the correct one above

Question

Simplify:

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

Answer

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 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.

Compare your answer with the correct one above

Question

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

Answer

Multiply both numberator and denominator by :

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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 :

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

Compare your answer with the correct one above

Question

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

Answer

First, divide 100 by as follows:

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

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

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

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

Now, divide this by :

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

Compare your answer with the correct one above

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