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Consider the complex number $$ \lambda = (1-2\mu)+2\mu\cos\theta+i\nu\sin \theta $$ where $i=\sqrt{-1}$, $\mu,\nu$ are constants, and $\mu>0$.


How can I get that $|\lambda|\leq 1$ for all $0\leq\theta\leq 2\pi$ if and only if $\nu^2\leq 2\mu\leq 1$?

What I have thought geometrically is that one can rewrite the formula as

$$ \lambda - (1-2\mu)=2\mu\cos\theta+i\nu\sin \theta $$ then $\lambda$ is on an ellipse on the complex plane, where the length of the axes are $2\mu$ and $\nu$. $\nu^2\leq 2\mu\leq 1$ implies $2\mu\leq 1$ and $\nu\leq 1$. I have no idea how to go on. An attempt to do it algebraically doesn't seem to work.

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Are you sure you have written the question (and the expected answer) carefully? Let $\theta=0$ and $\mu=\frac{1}{2}$. Then $\nu$ can take any values $<\infty$ and your expected conditions failed! – Tapu Dec 28 '11 at 8:31
up vote 4 down vote accepted

Geometrically, we want to find the values of the parameters $(\mu, \nu)$ such that the ellipse $$ \frac{(x - (1-2\mu))^2}{(2\mu)^2} + \frac{y^2}{\nu^2} = 1 \tag{E} $$ lies completely inside the unit circle $$ x^2 + y^2 = 1. \tag{C} $$ First of all, note that the left-most point of the ellipse -- namely, $(1-4\mu, 0)$ -- will lie inside the circle C if and only if $2 \mu \leqslant 1$. So we are left with finding the range of $\nu$ for a given $\mu$.

The ellipse will lie inside the unit circle if and only if, for all $x$ in the range $[1- 4\mu, 1]$, the height of the ellipse corresponding to the absicissa $x$ is smaller than or equal to the height of the circle: $$ \nu^2 \left(1 - \frac{(x - (1-2\mu))^2}{(2\mu)^2} \right) \leqslant 1 - x^2. \tag{$\ast$}$$

Here's a picture illustrating the final condition:

ellipses and circle

The picture shows the unit circle (in black), and three ellipses centered at $(1-2\mu, 0)$. The green ellipse is completely inside the circle, the red one exceeds it, and purple one is just inside the circle (touching it at $(1,0)$).

We pick an arbitrary “test point” $x$ is an arbitrary in the range $[1-4\mu, 1]$. The height of the ellipse at that point is denoted $y_e$ and that of the circle is denoted $y_c$. The inequality $(\ast)$ simply says that $y_e \leqslant y_c$ for all $x$ in the given range.

Our condition $(\ast)$ can be simplified as follows: $$ \begin{array}{crl} \iff& \nu^2 \left( 1 - \left( 1 - \frac{1-x}{2\mu} \right)^2 \right) &\leqslant 1 - x^2 \\ \iff & \nu^2 \left( \frac{1-x}{\mu} - \frac{(1-x)^2}{4 \mu^2} \right) &\leqslant 1 - x^2 \end{array} $$ Canceling a $1-x \geqslant 0$ factor, we have the equivalent condition $$ \frac{\nu^2}{\mu} - \frac{\nu^2(1-x)}{4 \mu^2} \leqslant 1 + x \tag{$\dagger$} $$ for all $x \in [1-4 \mu, 1]$.

  • Now in one direction, assuming $(\dagger)$ is satisfied, and plugging in $x = 1$ in it, we conclude that $\nu^2 \leqslant 2 \mu$.

  • In the other direction, we should prove that if $\nu^2 \leqslant 2 \mu \leqslant 1$, then $(\dagger)$ is satisfied for all $x \in [1-4\mu, 1]$. But this is straightforward: $$ \begin{align*} \frac{\nu^2}{\mu} - \frac{\nu^2(1-x)}{4 \mu^2} &\leqslant 2 - \frac{(1-x)}{2 \mu} \\ &\leqslant 2 - \frac{(1-x)}{1} \\ &= 1+x. \end{align*} $$

Thus we have showed that the ellipse lies inside the circle if and only if $\nu^2 \leqslant 2 \mu \leqslant 1$.

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Thanks for your answer. What tool did you use for drawing the picture? – Jack Dec 29 '11 at 3:18
@Jack You're welcome. The picture was drawn using GeoGebra. – Srivatsan Dec 29 '11 at 3:23

I am not able to solve the problem completely, but the following is what I can get so far. I want to share it so that it may lead to the complete solution eventually:

If $\lambda = (1-2\mu)+2\mu\cos\theta+i\nu\sin \theta$, then $$|\lambda|^2 = [(1-2\mu)+2\mu\cos\theta]^2+(\nu\sin \theta)^2$$ $$=(1-2\mu)^2+4\mu(1-2\mu)\cos\theta+4\mu^2\cos^2\theta+\nu^2\sin^2\theta,$$ hence, $|\lambda|^2\leq 1$ if and only if $$4\mu(\mu-1)+4\mu(1-2\mu)\cos\theta+4\mu^2\cos^2\theta+\nu^2\sin^2\theta\leq0.$$

If $\nu^2\geq 2\mu\geq 1$, then $$4\mu(\mu-1)+4\mu(1-2\mu)\cos\theta+4\mu^2\cos^2\theta+\nu^2\sin^2\theta$$ $$\geq4\mu(\mu-1)+4\mu(1-2\mu)\cos\theta+4\mu^2\cos^2\theta+2\mu\sin^2\theta$$ $$=4\mu(\mu-1)+4\mu(1-2\mu)\cos\theta+4\mu^2\cos^2\theta+2\mu(1-\cos^2\theta)$$ $$=2\mu(2\mu-1)+4\mu(1-2\mu)\cos\theta+2\mu(2\mu-1)\cos^2\theta$$ $$=2\mu(2\mu-1)(\cos\theta-1)^2\geq 0,$$ since $\mu>0$ by assumption.

If $\nu^2\leq 2\mu\leq 1$, then by above calculation, we have $$4\mu(\mu-1)+4(1-2\mu)\mu\cos\theta+4\mu^2\cos^2\theta+\nu^2\sin^2\theta$$ $$\leq2\mu(2\mu-1)(\cos\theta-1)^2\leq 0.$$

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As already said and since $\sin^2\theta=1-\cos^2\theta$, one has $1-|\lambda|^2=Q(\cos\theta)$ where $Q$ is the quadratic polynomial $$ Q(x)=4\mu(1-\mu)+4\mu(2\mu-1)x-4\mu^2x^2+\nu^2(x^2-1). $$ Thus, $|\lambda|^2\leqslant 1$ for every $\theta$ if and only if $Q(x)\geqslant0$ for every $|x|\leqslant1$. But $$ Q(x)=(1-x)A(x), $$ where the affine function $A$ is $A(x)=b-a(1+x)$ with $$ a=\nu^2-4\mu^2,\quad b=4\mu(1-2\mu). $$ One sees that $Q(x)\geqslant0$ for every $|x|\leqslant1$ if and only if $A(x)\geqslant0$ for every $|x|\leqslant1$ if and only if $A(1)=b-2a$ and $A(-1)=b$ are both nonnegative. The former condition reads $\nu^2\leqslant2\mu$ and, since $\mu\gt0$, the latter condition reads $2\mu\leqslant1$.

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Starting with Paul's answer, we have $|\lambda|^2\le1$

$$\Leftrightarrow 4\mu(\mu-1)+4\mu(1-2\mu)\cos\theta+4\mu^2\cos^2\theta+\nu^2\sin^2\theta\leq0$$

$$\Leftrightarrow (1-\cos\theta)\left[4\mu^2(1-\cos\theta)-4\mu+\nu^2(1+\cos\theta)\right]\leq0$$

$$\Leftrightarrow \underline{\cos\theta=1} \text{ or } \left[4\mu^2(1-\cos\theta)-4\mu+\nu^2(1+\cos\theta)\right]\leq0$$

The second factor reads $$(4\mu^2+\nu^2-4\mu)+\cos\theta(\nu^2-4\mu^2)\le0$$

$$\Leftrightarrow \cos\theta(4\mu^2-\nu^2)\ge(4\mu^2+\nu^2-4\mu)$$

Since this has to be satisfied for all $\theta$, taking the extreme values of $\cos\theta$, your claim follows.

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Sorry but I fail to see why you modified completely the previously deleted post above, and undeleted it. – Did Dec 28 '11 at 13:19
@DidierPiau, previous answer was wrong, while I think the current is correct :) – Tapu Dec 28 '11 at 15:25
Sure, the previous argument was incorrect, and you rightly deleted your post. The next move is the bizarre one. – Did Dec 28 '11 at 15:35
May be the time tag is the matter..? Sorry, I should have posted this as a new answer...right? – Tapu Dec 28 '11 at 18:27

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