Tag: Absolute Values

Compute the modulus and principal argument of given complex numbers

by RoRi

February 24, 2016

For each of the following complex numbers, compute the modulus and principal argument.

$2i$ .
$-3i$ .
$-1$ .
$1$ .
$-3 + \sqrt{3}i$ .
$\frac{1+i}{\sqrt{2}}$ .
$(-1+i)^3$ .
$(-1-i)^3$ .
$\frac{1}{1+i}$ .
$\frac{1}{(1+i)^2}$ .

First, the modulus is given by
$|2i| = \sqrt{2^2} = 2.$

The principal argument is the unique $\theta \in (-\pi, pi]$ such that

$x = r \cos \theta, \qquad y = r \sin \theta.$

In this case we have $x = 0$ , $y = 2$ and $r = \sqrt{2}$ so the principal argument satisfies

$2 \cos \theta = 0, \ \ 2 \sin \theta = 2 \qquad \implies \qquad \theta = \frac{\pi}{2}.$
First, the modulus is given by
$|-3i| = \sqrt{(-3)^2} = 3.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$3 \cos \theta = 0, \ \ 3 \sin \theta = -3 \qquad \implies \qquad \theta = -\frac{\pi}{2}.$
The modulus is given by
$|-1| = \sqrt{(-1)^2} = 1.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\cos \theta = -1, \ \ \sin \theta = 0 \qquad \implies \qquad \theta = \pi.$
The modulus is given by
$|1| = \sqrt{1^2} = 1.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\cos \theta = 1, \ \ \sin \theta = 0 \qquad \implies \qquad \theta = 0.$
The modulus is given by
$|-3+\sqrt{3}i| = \sqrt{(-3)^2 + \sqrt{3}^2} = \sqrt{12} = 2 \sqrt{3}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && 2 \sqrt{3} \cos \theta &= -3 & \text{and} && 2 \sqrt{3} \sin \theta &= \sqrt{3} \\ \implies && \cos \theta &= -\frac{\sqrt{3}}{2} & \text{and} && \sin \theta &= \frac{1}{2} \\ \implies && \theta &= \frac{5 \pi}{6}. \end{align*}$
The modulus is given by
$\left| \frac{1 + i}{\sqrt{2}} \right| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2} = \sqrt{1} = 1.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\cos \theta = \frac{1}{\sqrt{2}}, \ \ \sin \theta = \frac{1}{\sqrt{2}} \qquad \implies \qquad \theta = \frac{\pi}{4}.$
First, we write $(-1+i)^3$ in the form $a+bi$ ,
$(-1+i)^3 = 2 + 2i.$

Therefore, the modulus is

$|(-1+i)^3| = |2+2i| = \sqrt{2^2 + 2^2} = 2\sqrt{2}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && 2 \sqrt{2} \cos \theta &= 2 & \text{and} && 2 \sqrt{2} \sin \theta &= 2 \\ \implies && \cos \theta &= \frac{1}{\sqrt{2}} & \text{and} && \sin \theta &= \frac{1}{\sqrt{2}} \\ \implies && \theta &= \frac{\pi}{4}. \end{align*}$
First, we rewrite $(-1-i)^3$ in the form $a+bi$ ,
$(-1-i)^3 = (-1-i)(-1-i)(-1-i) = (2i)(-1-i) = 2 - 2i.$

Therefore, the modulus is

$|(-1-i)^3| = |2 - 2i| = \sqrt{2^2 + (-2)^2} = 2 \sqrt{2}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && 2 \sqrt{2} \cos \theta &= 2 & \text{and} && 2 \sqrt{2} \sin \theta &= -2 \\ \implies && \cos \theta &= \frac{1}{\sqrt{2}} & \text{and} && \sin \theta &= -\frac{1}{\sqrt{2}} \\ \implies && \theta &= -\frac{\pi}{4}. \end{align*}$
First, we rewrite $\frac{1}{1+i}$ in the form $a+bi$ ,
$\frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} = \frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i.$

Therefore, the modulus is

$\left| \frac{1}{1+i} \right| = \left|\frac{1}{2} - \frac{1}{2}i \right| = \sqrt{\left( \frac{1}{2} \right)^2 + \left(\frac{-1}{2}\right)^2} = \frac{1}{\sqrt{2}}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && \frac{1}{\sqrt{2}} \cos \theta &= \frac{1}{2} & \text{and} && \frac{1}{\sqrt{2}} \sin \theta &= -\frac{1}{2} \\ \implies && \cos \theta &= \frac{1}{\sqrt{2}} & \text{and} && \sin \theta &= -\frac{1}{\sqrt{2}} \\ \implies && \theta &= -\frac{\pi}{4}. \end{align*}$
First, we rewrite $\frac{1}{(1+i)^2}$ in the form $a+bi$ ,
$\frac{1}{(1+i)^2} = \frac{1}{2i} = -\frac{1}{2}i .$

The modulus is

$\left| \frac{1}{(1+i)^2} \right| = \left| \frac{-1}{2}i\right| = \sqrt{\left( \frac{-1}{2} \right)^2} = \frac{1}{2}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\frac{1}{2} \cos \theta = 0, \ \ \frac{1}{2} \sin \theta = -\frac{1}{2} \qquad \implies \qquad \theta = -\frac{\pi}{2}.$

Compute the absolute values of given complex numbers

by RoRi

February 24, 2016

For each of the following complex numbers, compute the absolute value.

$1+i$ .
$3 + 4i$ .
$\frac{1+i}{1-i}$ .
$1+i+i^2$ .
$i^7 + i^{10}$ .
$2(1-i) + 3(2+i)$ .

Using the formula for the absolute value of a complex number we compute,
$|1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}.$
Using the formula for the absolute value of a complex number we compute,
$|3+4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.$
Using the fact that $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$ and the formula for the absolute value of a complex number we have,
$\left| \frac{1+i}{1-i} \right| = \frac{|1+i|}{|1-i|} = \frac{\sqrt{2}}{\sqrt{2}} = 1.$
Using $i^2 = -1$ and the formula for the absolute value of a complex number we have
$|1 + i + i^2| = |i| = \sqrt{1^2} = 1.$
Using that $i^4 = 1$ we have
$|i^7 + i^{10}| = |i^3 + i^2| = |-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}.$
We compute,
$|2(1-i) + 3(2+i)| = |2 - 2i + 6 + 3i| = |8 + i| = \sqrt{8^2 + 1^2} = \sqrt{65}.$

Compute the integral from 0 to x of f(t) for the given functions

by RoRi

December 19, 2015

Find a formula to compute

$F(x) = \int_0^x f(t) \, dt$

for all $x \in \mathbb{R}$ for the following function $f(t)$ .

$\displaystyle{f(t) = (t + |t|)^2}$ .
The function,
$f(t) = \begin{cases} 1-t^2 & \text{if } |t| \leq 1, \\ 1 - |t| & \text{if } |t| > 1. \end{cases}$
$f(t) = e^{-|t|}$ .
$f(t) =$ the maximum of 1 and $t^2$ .

We know from this exercise (Section 5.5, Exercise #13) that
$F(x) = \int_0^x (t + |t|)^2 \, dt = \frac{2}{3} x^2 (x + |x|).$
If $\abs{x} \leq 1$ , then $f(t) = 1-t^2$ over the whole integral, and so
$F(x) = \int_0^x f(t) \, dt = \int_0^x (1-t^2) \, dt = \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^x = x - \frac{x^3}{3}.$

Then, if $x > 1$ we have

$\begin{align*} F(x) &= \int_0^x f(t) \, dt \\[9pt] &= \int_0^1 (1-t^2) \, dt + \int_1^x 1-t \, dt \\[9pt] &= \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^1 + \left( t - \frac{t^2}{2} \right) \Bigr \rvert_1^x \\[9pt] &= \frac{2}{3} + x - \frac{x^2}{2} - \frac{1}{2} \\[9pt] &= x - \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}. \end{align*}$

(Since $x > 1$ we have $\frac{|x|}{x} = 1$ so this equation works. This is the form Apostol wrote these answers as in the back of the book, so I’m getting our answers to match his. I wouldn’t have written them this way otherwise.)

Finally, if $x < -1$ we have

$\begin{align*} F(x) &= \int_0^x f(t) \, dt \\[9pt] &= \int_0^{-1} (1-t^2) \, dt + \int_{-1}^x (1+t) \, dt \\[9pt] &= \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^{-1} + \left( t + \frac{t^2}{2} \right) \Bigr \rvert_{-1}^x \\[9pt] &= -\frac{2}{3} + x + \frac{x^2}{2} + \frac{1}{2} \\[9pt] &= x + \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}. \end{align*}$

Since the formulas for $F(x)$ are the same for $x < -1$ and $x > 1$ are the same we have

$F(x) = x + \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}$

for $|x| > 1$ .
We consider two cases. If $x \geq 0$ then
$F(x) = \int_0^x e^{-|t|} \, dt = \int_0^x e^{-t} \, dt = -e^{-t} \Bigr \rvert_0^x = 1-e^{-x}.$

If $x < 0$ then

$F(x) = \int_0^x e^{-|t|} \, dt = \int_0^x e^t \, dt = e^x - 1.$
Since the maximum of 1 and $t^2$ is equal to 1 if $|t| \leq 1$ and is equal to $t^2$ if $|t| > 1$ we consider three cases ( $|x| \leq 1$ , $x > 1$ and $x < -1$ ).
For $|x| \leq 1$ we have

$F(x) = \int_0^x f(t) \, dt = \int_0^x dt = x.$

For $x > 1$ we have

$F(x) = \int_0^1 dt + \int_1^x t^2 \, dt = 1 + \frac{t^3}{3}\Bigr \rvert_1^x = \frac{x^3}{3} + \frac{2}{3} = \frac{x^3}{3} + \frac{|x|}{x} \frac{2}{3}$

For $x < -1$ we have

$F(x) = \int_0^{-1} dt + \int_{-1}^x t^2 \, dt = -1 + \frac{t^3}{3} \Bigr \rvert_{-1}^x = +\frac{x^3}{3} + \frac{|x|}{x} \frac{2}{3}.$

Find an integral formula involving the absolute value function

by RoRi

October 15, 2015

Prove that

$\int_0^x (t + |t|)^2 \, dt = \frac{2x^2}{3} (x + |x|)$

for all $x \in \mathbb{R}$ .

Proof. We consider two cases:
Case #1: If $x \geq 0$ then $t = |t|$ for all $t \in [0,x]$ so,

$\begin{align*} \int_0^x (t + |t|)^2 \, dt &= \int_0^x 4t^2 \, dt\\ &= \frac{4}{3} x^3 \\ &= \frac{2}{3} x^2 (2x) \\ &= \frac{2}{3} x^2 (x + |x|). \end{align*}$

The final equality follows since $x = |x|$ for $x \geq 0$ .
Case #2: If $x < 0$ . Then $|t| = -t$ for all $t \in [x,0]$ . So,

$\begin{align*} \int_0^x (t + |t|)^2 \, dt &= \int_0^x (t - t)^2 \, dt \\ &= 0 \\ &= \frac{2}{3} x^2 (x + |x|) \end{align*}$

since $x + |x| = 0$ for all $x < 0. \qquad \blacksquare$

Find the integral of the absolute value function

by RoRi

October 14, 2015

Show that

$\int_0^x |t| \, dt = \frac{1}{2} x |x|$

for all $x \in \mathbb{R}$ .

Proof. We consider three cases.
Case #1: If $x = 0$ , then both sides of the equation are 0, so the equation holds.
Case #2: If $x > 0$ , then $|t| = t$ for all $t \in [0,x]$ so,

$\int_0^x |t| \, dt = \int_0^x t \, dt = \frac{1}{2}x^2 = \frac{1}{2}x |x|.$

The last equality follows since $x > 0$ implies $x = |x|$ .
Case #3: If $x < 0$ , then $|t| = -t$ for all $\in [x,0]$ so,

$\begin{align*} \int_0^x |t| \, dt &= - \int_0^x t \, dt \\ &= \int_x^0 t \, dt \\ &= \left. \frac{1}{2}t^2 \right|_x^0 \\ &= -\frac{1}{2}x^2 \\ &= \frac{1}{2} x |x|. \end{align*}$

The final line follows since $|x| = -x$ since $x < 0. \qquad \blacksquare$

Prove some continuity properties of a function

by RoRi

August 31, 2015

Given a function $f$ and a closed interval $[a,b]$ such that

$|f(u) - f(v)| \leq |u-v|$

for all $u,v \in [a,b]$ .

Prove $f$ is continuous at every point $x \in [a,b]$ .
If $f$ is integrable on $[a,b]$ prove
$\left| \int_a^b f(x) \, dx - (b-a)f(a) \right| \leq \frac{(b-a)^2}{2}.$
If $c \in [a,b]$ is any point prove
$\left| \int_a^b f(x) \, dx - (b-a) f(c) \right| \leq \frac{(b-a)^2}{2}.$

Proof. Let $p$ be any point in $[a,b]$ . Then, for any $\varepsilon > 0$ let $\delta = \varepsilon$ . Then
$\begin{align*} |x-p| < \delta &&\implies && |x-p| &< \varepsilon \\ && \implies && |f(x) - f(p)| &< \varepsilon. \end{align*}$

Thus, for any $\varepsilon > 0$ , we have $|f(x) - f(p)| < \varepsilon$ whenever $|x-p| < \delta$ . Hence,

$\lim_{x \to p} f(x) = f(p).$

Therefore, $f$ is continuous at every point $p \in [a,b]. \qquad \blacksquare$
Proof. Since $f(a)$ is a constant (the value of the function $f$ evaluated at the constant $a$ ), we have
$(b-a) f(a) = \int_a^b f(a) \, dx.$

Therefore, we can compute,

$\begin{align*} \left| \int_a^b f(x) \, dx - (b-a) f(a) \right| &= \left| \int_a^b f(x) \, dx - \int_a^b f(a) \, dx \right| \\ &= \left| \int_a^b (f(x) - f(a)) \, dx \right|\\ &\leq \int_a^b |f(x) - f(a)| \, dx. \end{align*}$

The final inequality follows since, using the monotone property of the integral,

$\begin{align*} - |f(x)| \leq f(x) \leq |f(x)| && \implies && \int_a^b -|f(x)| \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b |f(x)| \, dx \\ && \implies && - \int_a^b |f(x)| \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b |f(x)| \, dx \\ && \implies && \left| \int_a^b f(x) \, dx \right| \leq \int_a^b |f(x)| \, dx. \end{align*}$

Then, since $|f(x) - f(a)| \leq |x-a|$ by our hypothesis on $f$ , we have

$\begin{align*} \int_a^b |f(x) - f(a)| \, dx &\leq \int_a^b |x-a| \, dx \\ &= \int_a^b (x-a) \, dx &(x-a\geq 0 \text{ since } x \in [a,b]) \\ &= \frac{b^2}{2} - \frac{a^2}{2} - ab + a^2 \\ &= \frac{b^2-2ab+a^2}{2} \\ & = \frac{(b-a)^2}{2}. \qquad \blacksquare \end{align*}$
Proof. The proof proceeds similarly to that of part (b),
$\begin{align*} \left| \int_a^b f(x) \, dx - (b-a) f(c) \right| &= \left| \int_a^b f(x) \, dx - \int_a^b f(c) \, dx \right|\\ &= \left| \int_a^b (f(x) - f(c)) \, dx \right| \\ &\leq \int_a^b | f(x) - f(c) | \, dx \\ &\leq \int_a^b |x-c| \, dx. \end{align*}$

Then, since $c \in [a,b]$ , we break the integral into two pieces (to deal with the fact that $|x-c| = -(x-c)$ for $x < c$ and $|x-c| = x-c$ for $x \geq c$ ):

$\begin{align*} \int_a^b |x-c| \, dx &= \int_a^c -(x-c) \, dx + \int_c^b (x-c) \, dx \\ &= \left. -\frac{x^2}{2} \right|_a^c + c \cdot x \biggr \rvert_a^c + \left. \frac{x^2}{2} \right|_c^b - c \cdot x \biggr \rvert_c^b \\ &= \frac{a^2-c^2}{2} + c(c-a) + \frac{b^2-c^2}{2} - c(b-c) \\ &= \frac{a^2 - 2c^2 + b^2}{2} + 2c^2-ac-bc \\ &= \frac{a^2+b^2}{2} + c^2 - c(b+a). \end{align*}$

But now, we know $c \in [a,b]$ implies $c \leq b$ , so we have the inequality,

$\begin{align*} \frac{a^2+b^2}{2} + c^2 - c(b+a) &= \frac{a^2+b^2}{2} + c(c-(b+a)) \\ &\leq \frac{b^2+a^2}{2} + b(b-b-a) \\ &= \frac{b^2+a^2}{2} - ab \\ &= \frac{b^2 - 2ab + a^2}{2} \\ &= \frac{(b-a)^2}{2}. \qquad \blacksquare \end{align*}$

Evaluate the limit

by RoRi

August 26, 2015

Evaluate

$\lim_{x \to 0^-} \frac{|x|}{x}.$

Since $|x| = -x$ for $x < 0$ , and the limit is taken as $x$ approaches 0 through negative values we have,

$\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1.$

Evaluate the limit

by RoRi

August 26, 2015

Evaluate

$\lim_{x \to 0^+} \frac{|x|}{x}.$

Since $|x| = x$ for $x > 0$ , and the limit is taken as $x$ approaches 0 through positive values, we have,

$\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1.$

Find the integral from -π to x of |1/2 + cos t|

by RoRi

August 4, 2015

For $0 \leq x \leq \pi$ , compute the following integral:

$\int_{-\pi}^x \left| \frac{1}{2} + \cos t \right| \, dt.$

Since we are taking the integral of an absolute value, first we want to determine on which intervals $\frac{1}{2} + \cos t$ is positive and on which it is negative. We have,

$\left| \frac{1}{2} + \cos t \right| = \begin{dcases} \frac{1}{2} + \cos t & \text{if } -\pi \leq x \leq -\frac{2 \pi}{3}, \text{ or } \frac{2 \pi}{3} \leq x \leq \pi \\ - \left( \frac{1}{2} + \cos t \right) & \text{if } -\frac{2 \pi}{3} \leq x \leq \frac{2 \pi}{3} \end{dcases}.$

Now must consider two cases. First, if $0 \leq x \leq \frac{2 \pi}{3}$ , then,

$\begin{align*} \int_{-\pi}^x \left| \frac{1}{2} + \cos t \right| \, dt &= - \int_{-\pi}^{- \frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \, dt + \int_{- \frac{2 \pi}{3}}^x \left( \frac{1}{2} + \cos t \right) \, dt \\ &= - \left( - \frac{\pi}{3} + \frac{\pi}{2} \right) - \left( \sin \left(-\frac{2 \pi}{3} \right) - \sin (- \pi) \right) + \left( \frac{x}{2} + \frac{\pi}{3} \right) + \left( \sin x - \sin \left(-\frac{2 \pi}{3} \right) \right) \\ &= \frac{\pi}{6} + \sqrt{3} + \frac{x}{2} + \sin x. \end{align*}$

In the other case, that $\frac{2 \pi}{3} \leq x \leq \pi$ , then,

$\begin{align*} \int_{-\pi}^x \left| \frac{1}{2} + \cos t \right| \, dt &= -\int_{-\pi}^{-\frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \, dt + \int_{- \frac{2 \pi}{3}}^{\frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \,dt - \int_{\frac{2 \pi}{3}}^x \left( \frac{1}{2} + \cos t \right) \, dt \\ &= \left( - \frac{\pi}{6} + \frac{\sqrt{3}}{2} \right) + \left( \frac{2 \pi}{3} + \sqrt{3} \right) - \left( \frac{x}{2} - \frac{\pi}{3} \right) - \left( \sin x - \frac{\sqrt{3}}{2} \right) \\ &= \frac{5 \pi}{6} + 2 \sqrt{3} - \frac{x}{2} - \sin x. \end{align*}$

Compute the integral from 0 to π of |(1/2) + cos t|

by RoRi

August 4, 2015

Compute the following integral:

$\int_0^{\pi} \left| \frac{1}{2} + \cos t \right| \, dt.$

First, we note that

$\left| \frac{1}{2} + \cos t \right| = \begin{dcases} \frac{1}{2} + \cos t & \text{if } 0 \leq t \leq \frac{2 \pi}{3} \\ -\left(\frac{1}{2} + \cos t\right) & \text{if } \frac{2 \pi}{3} < t \leq \pi. \end{dcases}$

So, then we can compute,

$\begin{align*} \int_0^{\pi} \left| \frac{1}{2} + \cos t \right| \, dt &= \int_0^{\frac{2 \pi}{3}} \left( \frac{1}{2} + \cos t \right) \, dt - \int_{\frac{2 \pi}{3}}^{\pi} \left( \frac{1}{2} + \cos t \right) \, dt \\ &= \int_0^{\frac{ 2 \pi}{3}} \frac{1}{2} \, dt + \int_0^{\frac{2 \pi}{3}} \cos t \, dt - \int_{\frac{2 \pi}{3}}^{\pi} \frac{1}{2} \, dt - \int_{\frac{2 \pi}{3}}^{\pi} \cos t \, dt \\ &= \left. \frac{t}{2} \right|_0^{\frac{2 \pi}{3}} + (\sin t)\biggr \rvert_0^{\frac{2 \pi}{3}} - \left. \frac{t}{2} \right|_{\frac{2 \pi}{3}}^{\pi} - (\sin t) \biggr \rvert_{\frac{2 \pi}{3}}^{\pi} \\ &= \frac{2 \pi}{6} + \sin \left( \frac{2 \pi}{3} \right) - \left( \frac{\pi}{2} - \frac{2 \pi}{6} \right) - \left( \sin \pi - \sin \left( \frac{2 \pi}{3} \right) \right) \\ &= \frac{\pi}{3} + \frac{\sqrt{3}}{2} - \frac{\pi}{6} + \frac{\sqrt{3}}{2} \\ &= \frac{\pi}{6} + \sqrt{3}. \end{align*}$