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# Absolute Value

## Definition

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

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

## Technique

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

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

### 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{align} \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{align}

### 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{align} 2-10a &\geq 0 \\ 2 & \geq 10a \\ \frac{1}{5} &\geq a. \end{align}

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

Therefore,

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

Note by Arron Kau
3 years, 4 months ago

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thank you very much :-) · 3 years, 3 months ago

In ending step ..note that to take modulus of imag eniry number is different than simple modulas · 2 years, 9 months ago

Excellent... · 3 years, 4 months ago

I didn't quite understand the problem having $$i$$. At one stage the $$i$$ disappears. Can you please explain? · 3 years, 4 months ago

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, $$|10 - \sqrt11|$$ = $$\sqrt{10^2 + \sqrt{11}^2}$$ · 3 years, 4 months ago

Didn't know that, thanks!! · 3 years, 4 months ago

$$i$$ is defined as $$\sqrt{-1}$$ i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers). · 3 years, 4 months ago

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... · 3 years, 4 months ago

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

Explanation More Exercises Solved ! · 3 years, 4 months ago