For a real number, the absolute value of $x$, denoted $\lvert x \rvert$, is defined as

$\left|x \right| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0. \end{cases}$

## Evaluate $\left| 3\left( 17-55 \right) \right|$.

$\begin{aligned} \left| 3\left( 17-55 \right) \right| &= \left| 3\left( -38 \right) \right| \\ &= \left| -114 \right| \\ &= 114 _\square \end{aligned}$

## If $x=\frac{5+\sqrt{11}i}{2}$ and $y=\frac{5 -\sqrt{11}i}{2}$, what is the value of $\left| x + 3y \right|^2$?

$\begin{aligned} \left| x + 3y \right|^2 &= \left| \frac{5+\sqrt{11}i}{2} + 3\left( \frac{5 -\sqrt{11}i}{2} \right) \right|^2 \\ &= \left| \frac{5+\sqrt{11}i + 15 -3\sqrt{11}i}{2} \right|^2 \\ &= \left| \frac{20-2\sqrt{11}i}{2} \right|^2 \\ &= \left| 10 - \sqrt{11}i \right|^2 \\ &= \left( \sqrt{10^2 + \sqrt{11}^2} \right)^2 \\ &= \sqrt{111}^2 \\ &= 111 _\square \end{aligned}$

## What is the absolute value of $2 - 10a$?

The answer depends on the value of $a$. From the definition above, we have

$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } 2-10a \geq 0 \\ -(2-10a), & \text{if } 2-10a < 0 \end{cases}$

Solving the inequality on the right, we obtain

$\begin{aligned} 2-10a &\geq 0 \\ 2 & \geq 10a \\ \frac{1}{5} &\geq a. \end{aligned}$

It follows that $2-10a \geq 0 \Longleftrightarrow a \leq \frac{1}{5}.$

Therefore,

$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } a \leq \frac{1}{5} \\ 10a - 2, & \text{if } a > \frac{1}{5} \end{cases} _\square$

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## Comments

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TopNewestthank you very much :-)

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I like this format of note, I think I'll Start Share So !

Explanation More Exercises Solved !

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I didn't quite understand the problem having $i$. At one stage the $i$ disappears. Can you please explain?

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It is because the absolute value of a complex number ($a + bi$) is defined as $\sqrt{a^2 + b^2}$ , i.e., $|a + bi|$ = $\sqrt{a^2 + b^2}$. In the above question, $|10 - \sqrt11|$ = $\sqrt{10^2 + \sqrt{11}^2}$

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Didn't know that, thanks!!

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I am not sure but as far as my mind think I can say this is because when the sign of absolute value was removed.Then sign was changed to positive & i(iota) has nothing to do when there is is postive root.So,It would have been removed...

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$i$ is defined as $\sqrt{-1}$ i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers).

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Excellent...

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In ending step ..note that to take modulus of imag eniry number is different than simple modulas

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