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How do I change $ z = 1- \sin (\alpha) + i \cos (\alpha) $ to polar? I got $r = (2(1-\sin(\alpha))^{\frac{1}{2}} $. I have problems with the exponential part. What should I do now?

Kaliche
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    The argument can be computed as $\arg z = \arctan\frac{\Im z}{\Re z}+n\pi$, so that should give you a start. The term $n\pi$ comes from the fact that $\arctan$ has a period of $\pi$, rather than $2\pi$ - so you also need to determine the $n$ that gives the correct argument within a $2\pi$ period. Hint: think about what $\arg z$ must be to place $z$ in the correct quadrant... – Tomas Aschan Apr 24 '13 at 22:46

4 Answers4

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We have \begin{eqnarray} z&=&1-\sin\alpha+i\cos\alpha=1+i(\cos\alpha+i\sin\alpha)=1+e^{i\pi/2}e^{i\alpha}=1+e^{i\frac{\pi+2\alpha}{2}}\\ &=&\left(e^{-i\frac{\pi+2\alpha}{4}}+e^{i\frac{\pi+2\alpha}{4}}\right)e^{i\frac{\pi+2\alpha}{4}}=2\cos\left(\frac{2\alpha+\pi}{4}\right)e^{i\frac{\pi+2\alpha}{4}}. \end{eqnarray}

For $\alpha \in \frac{\pi}{2}+2\pi\mathbb{Z}$ we have $z=0$.

For $\alpha \in (-\frac{3\pi}{2},\frac{\pi}{2})+4\pi\mathbb{Z}$ we have $|z|=2\cos\left(\frac{2\alpha+\pi}{4}\right),\ \arg z=\frac{\pi+2\alpha}{4}$.

For $\alpha \in (\frac{\pi}{2},\frac{5\pi}{2})+4\pi\mathbb{Z}$ we have $|z|=-2\cos\left(\frac{2\alpha+\pi}{4}\right),\ \arg z=\frac{5\pi+2\alpha}{4}$.

HorizonsMaths
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Angle is always $\displaystyle\arctan\left(\frac{Im}{Re}\right)$.

bob.sacamento
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$1-\sin\alpha\ge 0$

If $1-\sin\alpha=0,\cos\alpha=0, z=0+i\cdot0$

If $1-\sin\alpha> 0$

$1-\sin\alpha=(\cos\frac\alpha2-\sin \frac\alpha2)^2$ and

$\cos\alpha=\cos^2\frac\alpha2-\sin^2\frac\alpha2$

So, $z=(1-\sin\alpha)+i(\cos\alpha)=(\cos\frac\alpha2-\sin \frac\alpha2)^2+i(\cos^2\frac\alpha2-\sin^2\frac\alpha2)$

If $\cos\frac\alpha2-\sin \frac\alpha2>0,$ $z=(\cos\frac\alpha2-\sin \frac\alpha2)\left(\cos\frac\alpha2-\sin \frac\alpha2+i(\cos\frac\alpha2+\sin \frac\alpha2)\right)$ $=\sqrt2\cos(\frac\alpha2+\frac\pi4)\left(\sqrt2\cos(\frac\alpha2+\frac\pi4)+i\sqrt2\sin(\frac\alpha2+\frac\pi4)\right)$ $=2\cos(\frac\alpha2+\frac\pi4)\left(\cos(\frac\alpha2+\frac\pi4)+i\sin(\frac\alpha2+\frac\pi4)\right)$

If $\cos\frac\alpha2-\sin \frac\alpha2<0,$ $z=(\sin \frac\alpha2-\cos\frac\alpha2)\left(\sin \frac\alpha2-\cos\frac\alpha2-i(\cos\frac\alpha2+\sin \frac\alpha2)\right)$ $=-2\cos(\frac\alpha2+\frac\pi4)\left(-\cos(\frac\alpha2+\frac\pi4)-i\sin(\frac\alpha2+\frac\pi4)\right)$ $=2\cos(\frac\alpha2+\frac{5\pi}4)\left(\cos(\frac\alpha2+\frac{5\pi}4)+i\sin(\frac\alpha2+\frac{5\pi}4)\right)$ as $\cos(\pi+y)=-\cos y,\sin(\pi+y)=-\sin y$

Now, $\cos\frac\alpha2-\sin \frac\alpha2=\sqrt2\cos(\frac\pi4+\frac\alpha2)$ which will be $<0$ if $2n\pi+\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac{3\pi}2\iff 4n\pi+\frac{\pi}2<\alpha<4n\pi+\frac{5\pi}2$

Similarly, $\cos\frac\alpha2-\sin \frac\alpha2$ which will be $>0$ if $2n\pi-\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac\pi2\iff 4n\pi-\frac{3\pi}2<\alpha<4n\pi+\frac\pi2$

lab bhattacharjee
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$1-\sin\alpha\ge 0$

If $1-\sin\alpha=0,\cos\alpha=0, z=0+i\cdot0$

If $1-\sin\alpha> 0$

Let $1-\sin\alpha=r\cos t, \cos\alpha=r\sin t$ where $r> 0$

Squaring & adding we get $r^2=2(1-\sin\alpha)=2(\cos\frac\alpha2-\sin\frac\alpha2)^2=4\cos^2(\frac\pi4+\frac\alpha2)$

On division, $\frac{r\sin t}{r\cos t}=\frac{\cos \alpha}{1-\sin \alpha}$ $\implies \tan t=\frac{\cos^2\frac\alpha2-\sin^2\frac\alpha2}{(\cos\frac\alpha2-\sin^2\frac\alpha2)^2}$ $=\frac{\cos\frac\alpha2+\sin\frac\alpha2}{\cos\frac\alpha2-\sin\frac\alpha2}$ $=\frac{1+\tan \frac\alpha2}{1-\tan\frac\alpha2}=\tan(\frac\pi4+\frac\alpha2)$

$t=n\pi+\frac\pi4+\frac\alpha2$ where $n$ is any integer

If $n$ is odd$=2m+1$(say), $t=(2m+1)\pi+\frac\pi4+\frac\alpha2,\sin t=\sin\{(2m+1)\pi+\frac\pi4+\frac\alpha2\}=-\sin(\frac\pi4+\frac\alpha2)$ and $\cos t=\cos\{(2m+1)\pi+\frac\pi4+\frac\alpha2\}=-\cos(\frac\pi4+\frac\alpha2)$

If $n$ is even$=2m$(say), $t=2m\pi+\frac\pi4+\frac\alpha2,\sin t=\sin(2m\pi+\frac\pi4+\frac\alpha2)=\sin(\frac\pi4+\frac\alpha2)$ and $\cos t=\cos(2m\pi+\frac\pi4+\frac\alpha2)=\cos(\frac\pi4+\frac\alpha2)$

Now, $\cos(\frac\pi4+\frac\alpha2)$ will be $<0$ if $2n\pi+\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac{3\pi}2\iff 4n\pi+\frac{\pi}2<\alpha<4n\pi+\frac{5\pi}2$

Then $r=-2\cos(\frac\pi4+\frac\alpha2)$ Clearly $t=(2m+1)\pi+\frac\pi4+\frac\alpha2\equiv \pi+\frac\pi4+\frac\alpha2\pmod{2\pi}$

Similarly, $\cos(\frac\pi4+\frac\alpha2)$ will be $>0$ if $2n\pi-\frac\pi2<\frac\pi4+\frac\alpha2<2n\pi+\frac\pi2\iff 4n\pi-\frac{3\pi}2<\alpha<4n\pi+\frac\pi2$

Then, $r=2\cos(\frac\pi4+\frac\alpha2),$ Clearly $t=2m\pi+\frac\pi4+\frac\alpha2\equiv \frac\pi4+\frac\alpha2\pmod{2\pi}$

lab bhattacharjee
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