How to multiply complex numbers

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ACT Math › How to multiply complex numbers

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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 the following expression, leaving no complex numbers in the denominator.

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

$\frac{1+5i}{8}$

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

$\frac{-2+4i}{8}$

$\frac{8+ 3i}{32}$

$\frac{20-4i}{15}$

Explanation

Solving this problem requires eliminating the nonreal term of the denominator. Our best bet for this is to cancel the nonreal term out by using the conjugate of the denominator.

Remember that for all binomials , there exists a conjugate such that $(a+b)\cdot(a-b) = (a^2-b^2)$ .

This can also be applied to complex conjugates, which will eliminate the nonreal portion entirely (since )!

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

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

$\frac{(3+2i)(4+4i)}{(16-16i^2)} = \frac{(3+2i)(4+4i)}{32}$ Simplify. Note .

FOIL the numerator.

Combine and simplify.

$\frac{4+20i}{32} = \frac{1+5i}{8}$ Simplify the fraction.

Thus, $\frac{3+2i}{4-4i} = \frac{1+5i}{8}$ .

Simplify the following:

Explanation

Begin by treating this just like any normal case of FOIL. Notice that this is really the form of a difference of squares. Therefore, the distribution is very simple. Thus:

Now, recall that $i=\sqrt{-1}$ . Therefore, is . Based on this, we can simplify further:

Which of the following is equal to ?

Explanation

Remember that since $i = \sqrt{-1}{}$ , you know that is . Therefore, is or . This makes our question very easy.

is the same as or

Thus, we know that is the same as or .

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

Which of the following is equivalent to ?

$i^{23}$

$i^{1000}$

$i^{82}$

$i^{30}$

Explanation

When dealing with complex numbers, remember that $i = \sqrt{-1}$ .

If we square , we thus get $i^2 = (\sqrt{-1})^2 = -1$ .

Yet another exponent gives us $i^3 = i^2 \cdot i = -1 \cdot i = -i$ OR $-\sqrt{-1}$ .

But when we hit , we discover that $i^4 = i^2 \cdot i^2 = -1 \cdot -1 = 1$

Thus, we have a repeating pattern with powers of , with every 4 exponents repeating the pattern. This means any power of evenly divisible by 4 will equal 1, any power of divisible by 4 with a remainder of 1 will equal , and so on.

Thus, $i^{23}$

$\frac{23}{4} = 5 \frac{3}{4}$

Since the remainder is 3, we know that $i^{23} = i^3 = -i$ .

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

Distribute:

$\textup{None of these are correct}$

Explanation

This equation can be solved very similarly to a binomial like . Distribution takes place into both the real and nonreal terms inside the complex number, where applicable.

$(2\cdot4) + (2\cdot 7i)$

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

Distribute and solve:

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

Explanation

This problem can be solved very similarly to a binomial like .

$((5\cdot3)+ (5 \cdot i)) + ((2 \cdot 6) + (2\cdot 2i))$

Simplify the following:

Explanation

Begin this problem by doing a basic FOIL, treating just like any other variable. Thus, you know:

Recall that since $i=\sqrt{-1}$ , . Therefore, you can simplify further:

The solution of $\sqrt{x-3}>2$ is the set of all real numbers such that:

Explanation

Square both sides of the equation: $x-3=2^{2}$

Then Solve for x:

Therefore,

What is the product of and

Explanation

Multiplying complex numbers is like multiplying binomials, you have to use foil. The only difference is, when you multiply the two terms that have in the them you can simplify the to negative 1. Foil is first, outside, inside, last

First

$7i\cdot6=42i$

Outside:

$7i\cdot2i=14i^2=-14$

Inside

$-3\cdot6=-18$

Last

$-3\cdot2i=-6i$

Add them all up and you get

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