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If $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2(\cos\theta+i\sin\theta),$ $0<\theta<\pi$, then prove that $|\alpha-i|=|\beta-i|$

My Attempt $$ z^2-2e^{i\theta}z+1=0\implies\alpha+\beta=2e^{i\theta}\quad\&\quad\alpha\beta=1\\ \alpha,\beta=\frac{2e^{i\theta}\pm\sqrt{4e^{2i\theta}-4}}{2}=e^{i\theta}\pm\sqrt{e^{2i\theta}-1}=e^{i\theta}\pm\sqrt{\cos2\theta-1+i\sin2\theta}\\ =e^{i\theta}\pm\sqrt{-2\sin^2\theta+2i\sin\theta\cos\theta}=e^{i\theta}\pm\sqrt{2\sin\theta}.\sqrt{-\sin\theta+i\cos\theta}\\ =e^{i\theta}\pm\sqrt{2\sin\theta}.\sqrt{\cos(\tfrac{\pi}{2}+\theta)+i\sin(\tfrac{\pi}{2}+\theta)}\\ =e^{i\theta}\pm\sqrt{2\sin\theta}.\Big[{\cos(\tfrac{\pi}{4}+\tfrac{\theta}{2})+i\sin(\tfrac{\pi}{4}+\tfrac{\theta}{2})}\Big]\\ $$

As it was asked as a multiple choice question with the solution being one of the options, I think my attempt seems to be more complicated. So what is suggested to be the easiest way to find the solution ?. Can I use geometry ?

Sooraj S
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    $e^{i\theta}\ne\cos\theta+\sin\theta$. So, which do you want? – Angina Seng Feb 28 '19 at 06:42
  • @LordSharktheUnknown srry abt that ... edited op. its $e^{i\theta}$ – Sooraj S Feb 28 '19 at 06:47
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    Letting $\theta = \pi/2$, the equation becomes $z+1/z=2i$ with roots $\alpha,\beta=(1\pm\sqrt{2})i$. These do not satisfy $|\alpha-1|=|\beta-1|$. They do, however, satisfy $|\alpha-i|=|\beta-i|$. – FredH Feb 28 '19 at 08:25
  • @FredH thanx for the correction, but is it possible to prove it easily rather than verify as u showed ? – Sooraj S Feb 28 '19 at 09:46

2 Answers2

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This problem can be worked out using polar coordinates, but it seems to be a bit simpler to use rectangular coordinates.

That $|\alpha-i|=|\beta-i|$ follows from proving this claim: If $z$ is any root of $z+\frac{1}{z}=2e^{i\theta}$ with $0 \lt \theta \lt \pi$, then $|z-i| = \sqrt{2}$.

Let $z = x + yi$ and let $2e^{i\theta} = a + bi$, so $4 = a^2 + b^2$ and $b \gt 0$. The quantity $x^2 + y^2$ will appear often, so abbreviate it as $K$.

With these substitutions, $z+\dfrac{1}{z}=2e^{i\theta}$ becomes $x + yi + \dfrac{x - yi}{K} = a + bi$. Thus $a = x(1 + 1/K)$ and $b = y(1 - 1/K)$. Substituting these into $4 = a^2 + b^2$, we find $$ \begin{align} 4 &= x^2(1 + 2/K + 1/K^2) + y^2(1 - 2/K + 1/K^2)\\ &= (x^2 + y^2) + 2(x^2 - y^2)/K + (x^2 + y^2)/K^2\quad\\ &= K + 2(K - 2y^2)/K + 1/K.\\ \end{align} $$ After multiplying by $K$ and rearranging, this becomes $$ 4y^2 = K^2 - 2K + 1. $$ Clearly, $2y = \pm(K - 1)$, but which sign is correct? Since $b = y(1 - 1/K)$, we have $bK = y(K - 1)$. The left side is positive, so $y$ and $K - 1$ must have the same sign. Thus we have $2y = x^2 + y^2 - 1$, or $2 = x^2 + (y-1)^2$. This defines a circle of radius $\sqrt{2}$ centered at $(0,1)$. In complex terms, the distance from $z$ to $i$ is $\sqrt{2}$, as claimed.

FredH
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  • could u pls explain how u chose the +ve sign ? – Sooraj S Mar 07 '19 at 11:19
  • We are given $0\lt\theta\lt\pi$, so $b$ is positive (it is the imaginary part of $2e^{i\theta}$). $K = |z|^2$ and $z\ne0$, so $K$ is positive. Thus $bK$ is positive. $bK = y(K - 1)$, so $y(K-1)$ is positive. That means $y$ and $K-1$ have the same sign, which rules out $2y = -(K-1)$. – FredH Mar 07 '19 at 11:27
  • what i did was$$ |z+\frac{1}{z}|^2=|z|^2+\frac{1}{|z|^2}+\frac{2}{|z|^2}Re(z^2)=4\implies4|z|^2=|z|^4+1+2Re(z^2) \\implies(|z|-1)^2=4y^2\implies||z|-1|=2|y|\implies|z|-1=\pm2y$$ – Sooraj S Mar 07 '19 at 21:47
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$\alpha^2+1 = 2e^{i\theta}\alpha \Rightarrow \left|\alpha^2+1 \right| = 2\left|\alpha \right| \Rightarrow |\alpha+i||\alpha-i| = 2 |\alpha|$

Also $|\alpha+i|^2+|\alpha-i|^2 = 2 \left[|\alpha|^2+1 \right]$

Hence $\circ{1}: \left|\alpha+i \right|+\left|\alpha-i \right| = \sqrt 2\left(|\alpha|+1\right) $

and $\left|\left|\alpha+i \right| - \left|\alpha-i \right|\right|= \sqrt 2 \left||\alpha|-1\right| $

Now $z+\dfrac{1}{z} = 2 e^{i\theta} \Rightarrow \overline{z}+\dfrac{1}{\overline{z}} = 2 e^{-i\theta} $ and subtracting gives $\left(z-\overline{z}\right) \left(1-\dfrac{1}{|z|^2}\right)=4i\sin \theta$

Thus we have that $\Im(z) \left(|z|-1 \right)>0$

WLOG we choose $\alpha$ such that $|\alpha|>1$, since the roots are $\alpha, \dfrac{1}{\alpha}$ so that $\Im(z)>0$ and $|\alpha+i|> |\alpha-i|$

Hence we have $\circ{2}: \left|\alpha+i \right| - \left|\alpha-i \right| = \sqrt 2 \left(|\alpha|-1\right)$

Now dividing $\circ{1}$ and $\circ{2}$ and using componendo-dividendo we get

$\dfrac{|\alpha+i|}{|\alpha-i|} = |\alpha|$ or $|\alpha-i| = \left|1+\beta i \right|$ since the other root $\beta = \dfrac{1}{\alpha}$

and thus we get $|\alpha-i|=|\beta-i|$

Hari Shankar
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