- #1
EdMel
- 13
- 0
Homework Statement
37. Let ##z=\frac{i(1+is)}{1-is}## where ##s\epsilon\mathbb{R}##.
(a) Show that
$$\text{Arg}(z)=
\begin{cases}
\quad\frac{\pi}{2}+2\arctan s & \qquad \text{for}\quad s\leq1,\\
-\frac{3\pi}{2}+2\arctan s & \qquad\text{for}\quad s>1.
\end{cases}$$
Homework Equations
The formula for the argument of a complex number ##z## called Arg(##z##), with real part ##x## and imaginary part ##y##, can be given as
$$\text{Arg}(z)=
\begin{cases}
\arctan\left(\frac{y}{x}\right) & \text{if }x>0,\\
\arctan\left(\frac{y}{x}\right)+\pi & \text{if }x<0\text{ and }y\geq0,\\
\arctan\left(\frac{y}{x}\right)-\pi & \text{if }x<0\text{ and }y<0,\\
\qquad\quad\frac{\pi}{2} & \text{if }x=0\text{ and }y>0,\\
\qquad-\frac{\pi}{2} & \text{if }x=0\text{ and }y<0.
\end{cases}$$
The angle sum formula for the tangent function is
$$\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan \beta}{1\mp\tan\alpha\tan\beta}.$$
The Attempt at a Solution
NOTE: My questions about my solution are included at the end.
##\qquad##I started by 'realizing' the denominator of ##z##:
$$z=\frac{i(1+is)}{1-is}=\frac{i(1+is)}{1 is}\times\frac{1+is}{1+is}=\frac{i(1+2is+i^{2}s^{2})}{1-i^{2}s^{2}}
=\frac{i(1+2is-s^{2})}{1+s^{2}}=\frac{i+2i^{2}s-is^{2}}{1+s^{2}}=\frac{-2s}{1+s^{2}}+i\frac{1-s^{2}}{1+s^{2}},$$
so,
$$z=\frac{-2s}{1+s^{2}}+i\frac{1-s^{2}}{1+s^{2}}.$$
It can be shown that the modulus ##|z|=1##, so the set of ##z## forms the unit circle, but missing the point ##(-1,0)##,
as there is no value for ##s## such that ##\frac{1-s^{2}}{1+s^{2}}## equals -1. However, the limit of
##\frac{1-s^{2}}{1+s^{2}}## as ##s\rightarrow\pm\infty## is -1.
Then, letting ##\theta## be the argument of ##z## and based on the ##a+ib## form of ##z## above I made the following table:
\begin{array}{|c|c|c|c|}
\mathbf{s} & \mathbf{Re(z)=\frac{-2s}{1+s^{2}}} & \mathbf{Im(z)=\frac{1-s^{2}}{1+s^{2}}} &
\mathbf{\text{Quadrant $$z$$ lies in on Argand diagram}} \\
s>1 & Re(z)<0 & Im(z)<0 & 3 \\
s=1 & Re(z)<0 & Im(z)=0 & \text{on the real axis, point (-1,0)} \\
0<s<1 & Re(z)<0 & Im(z)>0 & 2 \\
s=0 & Re(z)=0 & Im(z)>0 & \text{on the imaginary axis, point (0,1)} \\
-1<s<0 & Re(z)>0 & Im(z)>0 & 1 \\
s=-1 & Re(z)>0 & Im(z)=0 & \text{on the real axis, point (1,0)} \\
s<-1 & Re(z)>0 & Im(z)<0 & 4
\end{array}
Then let the ##x## be the real part of ##z## and ##y## be the imaginary part, then ##\frac{y}{x}=\frac{1-s^2}{-2s}##, and we get
##\text{Arg}(z)=\theta=
\begin{cases}
\arctan\left(\frac{1-s^2}{-2s}\right) & \text{if }x>0, & \text{for }s<0, & \text{Case 1,} \\
\arctan\left(\frac{1-s^2}{-2s}\right)+\pi & \text{if }x<0\text{ and }y\geq0, & \text{for }0<s\leq1, & \text{Case 2,} \\
\arctan\left(\frac{1-s^2}{-2s}\right)-\pi & \text{if }x<0\text{ and }y<0, & \text{for }1<s, & \text{Case 3,} \\
\qquad\quad\frac{\pi}{2} & \text{if }x=0\text{ and }y>0, & \text{for }s=0, & \text{Case 4,} \\
\qquad-\frac{\pi}{2} & \text{if }x=0\text{ and }y<0, & \text{for no values of }s, & \text{Case 5}.
\end{cases}##
From now on a I shall refer to the 'given' formula for Arg(z) as that given in the question and the 'derived' formula as that gained from the relevant equation given in Section 2. The proof is split up into the five cases given above.
Case 1 : ##\mathbf{s<0}##
##\qquad##'derived'
For ##s<0## the derived formula gives ##\tan\theta=\frac{1-s^{2}}{-2s}## ...(1).
##\qquad##'given'
Then, that given from the question has
$$\theta=\frac{\pi}{2}+2\arctan s\Rightarrow\tan\left(\frac{\theta}{2}-\frac{\pi}{4}\right) =s,$$
and using the angle sum formula for the tangent we can write
$$s=\frac{\tan\left(\frac{\theta}{2}\right) \tan\left(\frac{\pi}{4}\right) }
{1+\tan\left(\frac{\theta}{2}\right) \tan\left(\frac{\pi}{4}\right) }
=\frac{\tan\left(\frac{\theta}{2}\right) -1}
{1+\tan\left(\frac{\theta}{2}\right) .1}
,$$
which can be rearranged to
$$\tan\left(\frac{\theta}{2}\right) =\frac{s+1}{1-s}\qquad\mathbf{...(2)}.$$
We can again use the angle sum for tangent and (2) to get
$$\tan(\theta)
=\frac{2\tan\left(\frac{\theta}{2}\right) }
{1+\tan^{2}\left(\frac{\theta}{2}\right) }
=\frac{2.\frac{s+1}{1-s}}
{1+\left(\frac{s+1}{1-s}\right) ^{2}}
,$$
which can be re-arranged to get
$$\tan(\theta)=\frac{1-s^{2}}{-2s},$$
which agrees with (1), and we have shown that the given formula for Arg(z) is correct for Case 1.
Case 2 : ##\mathbf{0<s\leq1}##
##\qquad##'derived'
For ##0<s\leq1## the derived formula gives
$$\theta=\arctan{\frac{1-s^2}{-2s}}+\pi\Rightarrow\tan\left(\theta-\pi\right) =\frac{1-s^2}{-2s}.$$
Use the angle sum formula for tangent to get
$$\frac{1-s^2}{-2s}=\tan\left(\theta-\pi\right) =\frac{\tan\theta-\tan\pi}{1+\tan\theta\tan\pi},$$
and as ##\tan\pi=0## we get
$$\frac{1-s^2}{-2s}=\frac{\tan\theta-0}{1+\tan\theta.0}\Rightarrow\tan\theta=\frac{1-s^2}{-2s}\qquad\mathbf{...(3)}$$
##\qquad##'given'
As per Case 1 the given formula leads to
$$\tan\theta=\frac{1-s^2}{-2s}$$
which agrees with (3) and the given formula is valid for Case 2.
Case 3 : ##\mathbf{1<s}##
##\qquad##'derived'
For ##1<s## the derived formula gives
$$\theta=\arctan{\frac{1-s^2}{-2s}}-\pi\Rightarrow\tan\left(\theta+\pi\right) =\frac{1-s^2}{-2s}.$$
Use the angle sum formula for tangent to get
$$\frac{1-s^2}{-2s}=\tan\left(\theta+\pi\right) =\frac{\tan\theta+\tan\pi}{1-\tan\theta\tan\pi},$$
and as ##\tan\pi=0## we get
$$\frac{1-s^2}{-2s}=\frac{\tan\theta+0}{1-\tan\theta.0}\Rightarrow \tan\theta=\frac{1-s^2}{-2s}\qquad\mathbf{...(4)}$$
##\qquad##'given'
For s>1 the given formula has
$$\theta
=-\frac{3\pi}{2}
+2\arctan{s}\Rightarrow\tan\left(\frac{\theta}{2}+\frac{3\pi}{4}\right) =s.$$
Then using the angle sum formula for tan we can write
$$s= \tan\left(\frac{\theta}{2}+\frac{3\pi}{4}\right)
=\frac{\tan\left(\frac{\theta}{2}\right) +\tan\left(\frac{3\pi}{4}\right) }
{1-\tan\left(\frac{\theta}{2}\right) \tan\left(\frac{3\pi}{4}\right) }
$$
and as ##\tan\left(\frac{3\pi}{4}\right) =-1## we get
$$s=
\frac{\tan\left(\frac{\theta}{2}\right) +(-1) }
{1-\tan\left(\frac{\theta}{2}\right) .(-1)}
\Rightarrow \tan\left(\frac{\theta}{2}\right) =\frac{s+1}{1-s},$$
which is (2) from Case 1, and use of the angle sum formula for tangent again leads to
$$\tan\theta=\frac{1-s^{2}}{-2s},$$
which is the same as (4) and the given formula is true for Case 3.
Case 4 : ##\mathbf{s=0}##
##\qquad##'derived'
For ##s=0## the derived formula gives
$$\theta=\frac{\pi}{2}.$$
##\qquad##'given'
For ##s=0## given formula has
$$\theta=\frac{\pi}{2}+2\arctan{s}\Rightarrow \theta=\frac{\pi}{2}+2\arctan{0}\Rightarrow\theta=\frac{\pi}{2}+2k\pi,~k\epsilon\mathbb{R}.$$
The only ##k## that gives ##-\pi<\theta\leq\pi## is ##k=0## so we have it that
$$\theta=\frac{\pi}{2}+2.0.\pi=\frac{\pi}{2},$$
which equals the derived value above, so the given formula is true for Case 4.
Case 5 is not applicable as ##\theta## never equals ##-\frac{\pi}{2}##.
Thus, the given formula for Arg(##z##) hase shown to be true for ##z=\frac{i(1+is)}{1-is}## by proving Cases 1 to 4. ##\Box##
My Questions:
1. Am I on the right track here?
2. My main concern is I have not started from first principles and derived the given formula. I have only shown that is equal to something derived from first principles.
3. Should I include something about the limiting behavior of the given formula for Arg(z) as it approaches ##\frac{\pi}{2}##, ##\pi## and ##-\frac{\pi}{2}##?
Thanks in advance.
