Imaginary Numbers

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Questions 1 - 10

Simplify the following complex number expression:

$(\sqrt{-6})(\sqrt{-4})(\sqrt{-3})$

$-6i\sqrt{2}$

$-6i\sqrt{3}$

$-4i\sqrt{3}$

$-4i\sqrt{2}$

$-6i\sqrt{5}$

Explanation

Begin by simplifying the radicals using complex numbers:

$(\sqrt{-6})(\sqrt{-4})(\sqrt{-3})$

$=(i\sqrt{6})(2i)(i\sqrt{3})$

Multiply the factors:

$=(2i^2\sqrt{6})(i\sqrt{3})$

$=(2i^3\sqrt{18})$

Simplify. Remember that is equivalent to .

$=-2i\sqrt{18}$

$=-6i\sqrt{2}$

Which of the following is equivalent to $\frac{3+i}{3-4i}$ , where $i=\sqrt{-1}$ ?

$\frac{1}{5}+\frac{3}{5}i$

$\frac{1}{5}-\frac{3}{5}i$

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

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

Explanation

We will need to simplify the rational expression by removing imaginary terms from the denominator. Often in such problems, we want to multiply the numerator and denominator by the conjugate of the denominator, which will usually eliminate the imaginary term from the denominator.

In this problem, the denominator is . Remember that, in general, the conjugate of the complex number is equal to , where a and b are both nonzero constants. Thus, the conjugate of is equal to .

We need to multiply both the numerator and denominator of the fraction $\frac{3+i}{3-4i}$ by .

$\frac{3+i}{3-4i}=\frac{3+i}{3-4i}\cdot\frac{3+4i}{3+4i}=\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}$

Next, we use the FOIL method to simplify both the numerator and denominator. The FOIL method of binomial multiplication requires us to mutiply together the first, outside, inner, and last terms of each binomial and then sum them.

$\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}=\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}$

Now, we can start simplifying.

$\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}=\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}$

Use the fact that $i^2=(\sqrt{-1})^2=-1$ .

$\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}=\frac{9+12i+3i+4(-1)}{9+12i-12i-16(-1)}$

$=\frac{9+15i-4}{9+16}=\frac{5+15i}{25}=\frac{5}{25}+\frac{15i}{25}$

$=\frac{1}{5}+\frac{3}{5}i$

The answer is $\frac{1}{5}+\frac{3}{5}i$ .

Simplify the following complex number expression:

$(\sqrt{-6})(\sqrt{-4})(\sqrt{-3})$

$-6i\sqrt{2}$

$-6i\sqrt{3}$

$-4i\sqrt{3}$

$-4i\sqrt{2}$

$-6i\sqrt{5}$

Explanation

Begin by simplifying the radicals using complex numbers:

$(\sqrt{-6})(\sqrt{-4})(\sqrt{-3})$

$=(i\sqrt{6})(2i)(i\sqrt{3})$

Multiply the factors:

$=(2i^2\sqrt{6})(i\sqrt{3})$

$=(2i^3\sqrt{18})$

Simplify. Remember that is equivalent to .

$=-2i\sqrt{18}$

$=-6i\sqrt{2}$

Which of the following is equivalent to $\frac{3+i}{3-4i}$ , where $i=\sqrt{-1}$ ?

$\frac{1}{5}+\frac{3}{5}i$

$\frac{1}{5}-\frac{3}{5}i$

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

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

Explanation

In this problem, the denominator is . Remember that, in general, the conjugate of the complex number is equal to , where a and b are both nonzero constants. Thus, the conjugate of is equal to .

We need to multiply both the numerator and denominator of the fraction $\frac{3+i}{3-4i}$ by .

$\frac{3+i}{3-4i}=\frac{3+i}{3-4i}\cdot\frac{3+4i}{3+4i}=\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}$

$\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}=\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}$

Now, we can start simplifying.

$\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}=\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}$

Use the fact that $i^2=(\sqrt{-1})^2=-1$ .

$\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}=\frac{9+12i+3i+4(-1)}{9+12i-12i-16(-1)}$

$=\frac{9+15i-4}{9+16}=\frac{5+15i}{25}=\frac{5}{25}+\frac{15i}{25}$

$=\frac{1}{5}+\frac{3}{5}i$

The answer is $\frac{1}{5}+\frac{3}{5}i$ .

Simplify the following complex number expression: