Pls share ur ideas i wish to knw were i have gone wrong or is the text answr is wrong pls help(Headbang)(Headbang)(Headbang)
Printable View
Pls share ur ideas i wish to knw were i have gone wrong or is the text answr is wrong pls help(Headbang)(Headbang)(Headbang)
It helps if ALL of your PDF can be read. Is your original complex number $\displaystyle \displaystyle z = \cos{(2\theta)} + i\sin{(2\theta)}$?
yes yes
I agree with your evaluation of $\displaystyle \displaystyle \begin{align*} |1 - z| \end{align*}$. As for $\displaystyle \displaystyle \begin{align*} \arg{ (1 - z)} \end{align*}$ you will need to consider nine separate cases...
Case 1: $\displaystyle \displaystyle \begin{align*} 1 - z \end{align*}$ is in the first quadrant, so $\displaystyle \displaystyle \begin{align*} 1 - \cos{(2\theta)} > 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} -\sin{(2\theta)} > 0 \end{align*}$. The first is true for all $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$ except the integer multiples of $\displaystyle \displaystyle \begin{align*} \pi \end{align*}$, and the second is true in the region $\displaystyle \displaystyle \begin{align*} -\frac{\pi}{2} < \theta < 0 \end{align*}$ and the addition of all integer multiples of $\displaystyle \displaystyle \begin{align*} 2\pi \end{align*}$ for this region. Therefore, $\displaystyle \displaystyle \begin{align*} 1 - z \end{align*}$ is in the first quadrant when $\displaystyle \displaystyle \begin{align*} -\frac{\pi}{2} < \theta < 0 \end{align*}$ or any integer multiple of $\displaystyle \displaystyle \begin{align*} 2\pi \end{align*}$ being added to this region.
When we have a $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$ in this domain, then
$\displaystyle \displaystyle \begin{align*} \arg{(1 - z)} &= \arctan{ \left[ \frac{1 - \cos{(2\theta)}}{-\sin{(\theta)}} \right] } \\ &= \arctan{ \left[ \frac{\cos{(2\theta)} - 1}{\sin{(2\theta)}} \right] } \\ &= \arctan{ \left[ \frac{\cos^2{(\theta)} - \sin^2{(\theta)} - 1}{2\sin{(\theta)}\cos{(\theta)}} \right] } \\ &= \arctan{ \left[ \frac{-2\sin^2{(\theta)}}{2\sin{(\theta)}\cos{(\theta)}} \right] } \\ &= \arctan{ \left[ -\tan{(\theta)} \right] } \\ &= \arctan{ \left[ \tan{(-\theta)} \right] } \\ &= -\theta \end{align*}$
Case 2: $\displaystyle \displaystyle \begin{align*} 1 - z \end{align*}$ is in the second quadrant, so $\displaystyle \displaystyle \begin{align*} 1 - \cos{(2\theta)} < 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} -\sin{(2\theta)} > 0 \end{align*}$. The first of these is never true, so your complex number can never be in the second quadrant.
Case 3: $\displaystyle \displaystyle \begin{align*} 1 - z \end{align*}$ is in the third quadrant, so $\displaystyle \displaystyle \begin{align*} 1 - \cos{(2\theta)} < 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} -\sin{(2\theta)} < 0 \end{align*}$. The first of these is never true, so your complex number can never be in the third quadrant.
Case 4: $\displaystyle \displaystyle \begin{align*} 1 - z \end{align*}$ is in the fourth quadrant, so $\displaystyle \displaystyle \begin{align*} 1 - \cos{(2\theta)} > 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} -\sin{(2\theta)} < 0 \end{align*}$. The first of these is true for all $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$ except the integer multiples of $\displaystyle \displaystyle \begin{align*} \pi \end{align*}$, and the second is true when $\displaystyle \displaystyle \begin{align*} 0 < \theta < \frac{\pi}{2} \end{align*}$ or the addition of any multiples of $\displaystyle \displaystyle \begin{align*} 2\pi \end{align*}$ to this region. So when $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$ satisfies this case, then
$\displaystyle \displaystyle \begin{align*} \arg{(1 - z)} &= -\arctan{ \left[ \frac{1 - \cos{(2\theta)}}{-\sin{(2\theta)}} \right] } \\ &= -\left( -\theta \right) \\ &= \theta \end{align*}$
The other cases are when your complex number lies on the positive real axis, the negative real axis, the positive imaginary axis, the negative imaginary axis, and the origin. See how you go with the rest of them :)
Thank u,but wen z=x+iy ,arg z = tan inverse(y/x),then arg of 1-z= -sin2(theta)/1-cos2(theta) bt u did the prb other way round am i mistaken
my main question is even by thinking ur way arg (1-z)=pi/2 +theta ,since case1 is 0<theta<pi ,it wont lie in secnd quadrant so has to lie in first quadrant,but answer in text says arg(1-z)=theta- pi/2
You must understand the following:Quote:
Originally Posted by spacemenon
Suppose that $\displaystyle x\cdot y\ne 0 $ then
$\displaystyle Arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. $
No, I did this while I was half asleep. You will need to make the necessary changes. The simplification is very similar though.Quote:
Originally Posted by spacemenon
i am sorry bt u didnt answer this part of the question, u wrote
arg 1-z = arctan(1-cos2theta/sin2theta) instead of arg 1-z= arctan(-sin2(theta)/1-cos2(theta) ) why?
sry u already answered that i didnt see the post
so arg(1-z) =pi/2 +theta and mod(1-z)= 2sin (theta)
so according to ur theory y coordinatte is -sin2(theta) and x co0rdinate is 1-cos (2theta)
so case 1: 0<theta< pi
the complex no shld lie in 2nd quadrant as first quadrant -sin2theta is nt positive so only possibilty is second quadrant ,therfore we conclude x<0 and y >0 so arg(1-z) = pi/2+theta -pi =-pi/2+theta
