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Please help with a question that I am working on just now...:)

If $z=2e^{i\theta}$ where $0<\theta<\pi$, how can I find the real and imaginary parts of $w=(z-2)/(z+2)$? Hence, how can I calculate w and deduce that this complex number always lies parallel to the real axis in the Argand plane?

The following information can be used:

$\cos(2\theta)=2\cos^2(\theta)-1$ and

$\sin(2\theta)=2\cos(\theta)\sin(\theta)$

Thanks for all your help!!

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  • $\begingroup$ You might want to check the result you are asking to show in some simple case, say, when $z=2i$. $\endgroup$
    – Did
    Jan 18, 2015 at 18:30

3 Answers 3

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you can argue this geometrically as follows: i will work with $0 \le \theta \le \pi$ for $\theta$ in the lower half plane replace $\theta$ by $-\theta$

$z = 2e^{i\theta}$ lies on the circle with diameter $-2$ and $2$. that is the center of the circle is $0$ and the radius is $2$ now observe that the angle subtended by the diameter on the circle is $\pi/2$ this translates as $z - 2$ and $z + 2$ are orthogonal. we can say more $$z - 2 = 2 \sin(\theta/2)e^{i(\pi+\theta)/2}, \qquad z + 2 = 2\cos (\theta/2)e^{i\theta/2},$$ hence $$ w = \dfrac{z-2}{z+2} = i\tan (\theta/2) $$ so $w$ is a pure imaginary number therefore is parallel to imaginary axis.

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    $\begingroup$ You might want to check your final result in some simple case, say, when $z=2i$. $\endgroup$
    – Did
    Jan 18, 2015 at 18:29
  • $\begingroup$ @Did, thanks for catching the error. hopefully there is no more. $\endgroup$
    – abel
    Jan 18, 2015 at 18:48
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The simplest way consists in replacing $z$ with its value, then multiplying numerator and denominator by $\mathrm e^{-\mathrm i \frac\theta2}$ and using Euler's formulae.

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  • $\begingroup$ Ah, I see! That's really neat! I got the real part=0 and imaginary part to be tan(theta/2) $\endgroup$ Jan 18, 2015 at 18:03
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    $\begingroup$ That's right. It's the same method that allows to find a closed form for the sums $1+\cos\theta+\cos2\theta+\dots+\cos n\theta $ and $\sin\theta+\sin2\theta+\dots+\sin n\theta $. $\endgroup$
    – Bernard
    Jan 18, 2015 at 18:11
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it is perhaps worthwhile to observe in passing that there is an elementary geometric interpretation of these facts.

the points $P=2, M=-2$ and $Z=2e^{i\theta}$ all lie on the circle with radius $2$ centered at $O$.

the geometrical result that the angle subtended by a diameter ($MP$) at a point on the circumference ($Z$) is a right angle shows that the lines $ZP$ and $ZM$ are orthogonal, hence the ratio of $2z-2$ to $2z+2 = 2z-(-2)$ is a purely imaginary complex number. also its magnitude is $\frac{|ZP|}{|ZM|} = \tan Z \hat MP=\tan \frac{\theta}2$ by another theorem of geometry, that the angle subtended by a chord at the circumference is a half of the angle subtended by the same chord at the center (if both angles are on the same side of the chord)

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