Number complex (1 Viewer)

Abir dz · Feb 6, 2016

Hello

i want to transforme this ( z=1-cosx+isinx ) to polar form

And Thank you

InteGrand · Feb 6, 2016

Abir dz said:
Hello

i want to transforme this ( z=1-cosx+isinx ) to polar form

And Thank you

$\noindent We'll assume $x \in \left[0,2\pi\right)$ (if it's not, convert it to the equivalent angle in this range). We have$

$$\begin{align*} |z|^2 &= (1-\cos x)^2 +\sin^2 x \\ &= 1-2\cos x + \cos^2x +\sin^2 x \\ &= 2-2\cos x \quad (\text{using the Pythagorean trig. identity}) \\ &= 4 \left(\frac{1}{2} -\frac{1}{2}\cos x\right) \\ &= 4\sin^2 \frac{x}{2} \quad (\text{half-angle identity}).\end{align*}$

$\noindent Taking square roots, we have $|z| = 2\left| \sin \frac{x}{2}\right|$.$

$\noindent Now we'll find the principal argument (call it $\theta$). Note that if $x \neq 0$, $\mathrm{Re} (z) =1-\cos x >0$. For $x\neq 0$, note that$

$\begin{align*} \frac{\mathrm{Im}(z)}{\mathrm{Re}(z)} &= \frac{\sin x}{1-\cos x}\\ &= \cot \frac{x}{2} \quad (\text{this can be seen by using the } t\text{- formulae}) \\ &= \tan \left(\frac{\pi}{2} -\frac{x}{2}\right) \quad (\text{trig. identity}).\end{align*}$

$\noindent Since for any $x \in (0,2\pi)$, $\frac{\pi}{2}-\frac{x}{2} $ lies in the range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, taking inverse tangents of both sides above tells us that for $x\neq 0$, we have $\theta = \frac{\pi}{2} -x$. Note that if we substitute $x=0$, then $z$ is just 0, so the following formula holds for any $x\in [0, 2\pi)$: $z = 2\left| \sin \frac{x}{2}\right| \, \mathrm{cis} \left(\frac{\pi}{2} -x\right)$.$

Abir dz · Feb 6, 2016

Thank you a lot for this explanation

but you forgot (i)

i said : z= 1-cosx+ i sinx

not z= 1-cox + sinx

InteGrand · Feb 6, 2016

Abir dz said:
Thank you a lot for this explanation

but you forgot (i)

i said : z= 1-cosx+ i sinx

not z= 1-cox + sinx

$\noindent What do you mean? I did do it assuming $z= (1-\cos x)+i\sin x$.$

Abir dz · Feb 6, 2016

InteGrand said:
$\noindent What do you mean? I did do it assuming $z= (1-\cos x)+i\sin x$.$

yeah now i understand it thank you again

you Calculated the /z/² and i didn't attention for it

( sorry about my broken english )

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Number complex (1 Viewer)

Abir dz

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InteGrand

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Abir dz

New Member

InteGrand

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Abir dz

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