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Write the complex number $z=1 + \cos \alpha +i \sin \alpha$ in polar form

Here is what I got so far,

$r= |z|=((1+\cos \alpha)^2 +\sin^2 \alpha)^{1/2}$

this is not pretty number but I can simplify it a little bit, but what abour $arg(z)$

$arg(z)=arctan (\frac {\sin \alpha}{1+ \cos \alpha}$)

I'm not sure if this make sense to put into the polar form.

I also tried to break this in to 2 parts

$z=z_1 +z_2 $ where $z_1 =1=1(\cos 0 +i \sin 0)$ and $z_2= 1(\cos \alpha +i \sin \alpha)$ but that doesn't get me any where. Any help will be much appreciated.

enter image description here

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    $\begingroup$ HINT: First write this in standard form as $z=\cos(\alpha)+i(1+\sin(\alpha))$ $\endgroup$
    – Mufasa
    Jan 17, 2015 at 14:02
  • $\begingroup$ Do you mean $(1 + \cos \alpha) + i \sin \alpha$ (as your computations suggest)? If so, you can simplify your expressions for $r$ and $\arg z$ using half-angle identities. $\endgroup$ Jan 17, 2015 at 14:10
  • $\begingroup$ It looks like you've added 1 instead of i. $r^2=2(1+\sin\alpha)$, $\tan arg(z)=\frac{1+\sin\alpha}{\cos\alpha} = \tan\alpha + \sec\alpha$ $\endgroup$
    – PM 2Ring
    Jan 17, 2015 at 14:12
  • $\begingroup$ yes, I copied the question incorrectly. Sorry. $\endgroup$ Jan 17, 2015 at 14:18

2 Answers 2

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We have $1+\cos\alpha+i(\sin\alpha)$

so, arg$(z)=\arctan\dfrac{\sin\alpha}{1+\cos\alpha}$

If $1+\cos\alpha=\sin\alpha=0\iff\alpha=(2n+1)\pi,$ where $n$ is any integer

arg$(z)$ will be undefined

Otherwise, arg$(z)=\arctan\dfrac{2\cos\dfrac\alpha2\sin\dfrac\alpha2}{2\cos^2\dfrac\alpha2}=\arctan\left[\tan\left(\dfrac\alpha2\right)\right]$

Now use the definition

and adjust $\dfrac\alpha2$ to $\beta=m\pi+\dfrac\alpha2$ for some integer $m$ so that $\beta$ lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$

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  • $\begingroup$ I copied the question incorrectly, it's $1$ not $i$ $\endgroup$ Jan 17, 2015 at 14:17
  • $\begingroup$ @DianeVanderwaif, Adjusted $\endgroup$ Jan 17, 2015 at 14:18
  • $\begingroup$ so you got $arg(z)=\beta$ but I'm not sure I understand how you got $ arg(z)=\arctan\dfrac{2\cos\dfrac\alpha2\sin\dfrac\alpha2}{2\cos^2\dfrac\alpha2}$? $\endgroup$ Jan 17, 2015 at 14:26
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    $\begingroup$ @DianeVanderwaif, See mathworld.wolfram.com/Double-AngleFormulas.html $\endgroup$ Jan 17, 2015 at 14:27
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    $\begingroup$ @DianeVanderwaif: Another way to do this reduction is to use the tangent half-angle formulas: $\sin x = \frac{2t}{1 + t^2}, cos x = \frac{1 - t^2}{1 + t^2}, \tan x = \frac{2t}{1 - t^2}$. But it's probably simplest to just draw a diagram & solve it geometrically, as suggested by abel. $\endgroup$
    – PM 2Ring
    Jan 18, 2015 at 3:16
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i will use little bit geometry to get at the answer. let us keep $-\pi < \alpha < \pi$

let the point $o = 0, a = 1, u = \cos \alpha + i \sin \alpha$ and the point $z = 1 + u$ observe that $|u| = 1$ and the triangle $oaz$ is an isosceles triangle so the angle $aoz = \dfrac{\alpha}{2}$ and the side $|z| = 2\sin(\alpha/2)$
therefore in polar form $$ z = 2 \sin(\alpha/2) e^{i\alpha/2} \text{ if } 0 \le \alpha < \pi $$

$$ z = 2 \sin(-\alpha/2) e^{i\alpha/2} \text{ if } -\pi \le \alpha < 0 $$

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  • $\begingroup$ For some reason, my professor doesn't want $\alpha \in [-\pi,\pi]$ but he prefers $\alpha \in [0,2\pi]$, does this change anything? I can see that $u$ is all the point that lies on the circle of center is the origin and radius $1$, but I can't see how the angle $aoz = \frac{\alpha}{2}$ Can you explain this for me please. $\endgroup$ Jan 29, 2015 at 21:36
  • $\begingroup$ @XiaoXiaoZhen, if you need $\alpha \in [0,2\pi),$ then use $z = 2\sin(\alpha/2)e^{i(\pi+ \alpha/2)}$. another way to see is draw the unit circle and the isosceles triangle $oaz$ and mark the angles. i am sorry i don't know how to upload a figure. $\endgroup$
    – abel
    Jan 29, 2015 at 22:44
  • $\begingroup$ I'm not sure if I understand abel's idea correctly, but I just try my best to put together a picture, not very accurate but something like that, hopefully this will help you, Zhen. $\endgroup$ Jan 29, 2015 at 22:52
  • $\begingroup$ @abel from your answer $|z|=2\sin(\alpha/2)$ implies that $u$ and $a$ are perpendicular. But I don't see that in Diane's picture, and I'm not sure why $u$ has to be perpendicular to $a$. $\endgroup$ Jan 30, 2015 at 20:45

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