# complex number in polar form argument and modulus doubt

• Jul 9th 2013, 06:49 AM
spacemenon
complex number in polar form argument and modulus doubt
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)
• Jul 9th 2013, 06:53 AM
Prove It
Re: complex number in polar form argument and modulus doubt
It helps if ALL of your PDF can be read. Is your original complex number $\displaystyle z = \cos{(2\theta)} + i\sin{(2\theta)}$?
• Jul 9th 2013, 07:13 AM
spacemenon
Re: complex number in polar form argument and modulus doubt
yes yes
• Jul 9th 2013, 08:05 AM
Prove It
Re: complex number in polar form argument and modulus doubt
I agree with your evaluation of \displaystyle \begin{align*} |1 - z| \end{align*}. As for \displaystyle \begin{align*} \arg{ (1 - z)} \end{align*} you will need to consider nine separate cases...

Case 1: \displaystyle \begin{align*} 1 - z \end{align*} is in the first quadrant, so \displaystyle \begin{align*} 1 - \cos{(2\theta)} > 0 \end{align*} and \displaystyle \begin{align*} -\sin{(2\theta)} > 0 \end{align*}. The first is true for all \displaystyle \begin{align*} \theta \end{align*} except the integer multiples of \displaystyle \begin{align*} \pi \end{align*}, and the second is true in the region \displaystyle \begin{align*} -\frac{\pi}{2} < \theta < 0 \end{align*} and the addition of all integer multiples of \displaystyle \begin{align*} 2\pi \end{align*} for this region. Therefore, \displaystyle \begin{align*} 1 - z \end{align*} is in the first quadrant when \displaystyle \begin{align*} -\frac{\pi}{2} < \theta < 0 \end{align*} or any integer multiple of \displaystyle \begin{align*} 2\pi \end{align*} being added to this region.

When we have a \displaystyle \begin{align*} \theta \end{align*} in this domain, then

\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 \begin{align*} 1 - z \end{align*} is in the second quadrant, so \displaystyle \begin{align*} 1 - \cos{(2\theta)} < 0 \end{align*} and \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 \begin{align*} 1 - z \end{align*} is in the third quadrant, so \displaystyle \begin{align*} 1 - \cos{(2\theta)} < 0 \end{align*} and \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 \begin{align*} 1 - z \end{align*} is in the fourth quadrant, so \displaystyle \begin{align*} 1 - \cos{(2\theta)} > 0 \end{align*} and \displaystyle \begin{align*} -\sin{(2\theta)} < 0 \end{align*}. The first of these is true for all \displaystyle \begin{align*} \theta \end{align*} except the integer multiples of \displaystyle \begin{align*} \pi \end{align*}, and the second is true when \displaystyle \begin{align*} 0 < \theta < \frac{\pi}{2} \end{align*} or the addition of any multiples of \displaystyle \begin{align*} 2\pi \end{align*} to this region. So when \displaystyle \begin{align*} \theta \end{align*} satisfies this case, then

\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 :)
• Jul 9th 2013, 05:49 PM
spacemenon
Re: complex number in polar form argument and modulus doubt
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
• Jul 9th 2013, 05:55 PM
spacemenon
Re: complex number in polar form argument and modulus doubt
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
• Jul 9th 2013, 06:30 PM
Plato
Re: complex number in polar form argument and modulus doubt
Quote:

Originally Posted by spacemenon
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:
Suppose that $x\cdot y\ne 0$ then
$Arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.$
• Jul 9th 2013, 07:17 PM
Prove It
Re: complex number in polar form argument and modulus doubt
Quote:

Originally Posted by spacemenon
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

No, I did this while I was half asleep. You will need to make the necessary changes. The simplification is very similar though.
• Jul 9th 2013, 07:21 PM
spacemenon
Re: complex number in polar form argument and modulus doubt
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?
• Jul 9th 2013, 07:23 PM
spacemenon
Re: complex number in polar form argument and modulus doubt