$z^2+bz+1=0$, where $|b|\le2$. For which values of $b$ do the following hold?

Question

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This is to check whether my answer is correct or not.
I got for (i) $ $ $0<b\le2$, $ $ and for (ii) $\pm \frac{1+\pi^2}{\pi}$.

I’m not convinced at all about the method I used, so I can hardly tell.

I start by finding the two possible roots expressed in terms of $b$, by completing the square of the quadratic expression.
Now, if (i) is to be true then $z$ must be negative. So, I check the possible values for $b$ so that $z$ is negative. For (ii), I use Euler’s Identity, so $z$ must equal $\pi$. I solve the expression for $z$ in terms of $b$ when it equals $\pi$ to get the answer.

EDIT 1

I realized I missed a minus sign. Corrected (ii). Still unsure whether it's correct.

$ $

EDIT 2

I realized I missed another minus sign... corrected (ii). Still, it appears from the answers below that this is wrong. I can't see why.

I used Euler's Identity, and observed that $z=\pmπ$ if (ii) is true. I solved $\pm \pi=-\frac{b}{2}\pm\sqrt{\frac{b^2}{4}-1}$, to get a value for $b$.

Why is this method not giving the correct answer?

$ $

EDIT 3

I've got it! (my answer is below)

Answer 1

One possible method:

WLOG $b=2\cos y$ where $0\le y\le\pi$

$$\implies z=\cos y\pm i\sin y$$

$$1>|e^z|=e^{\cos y}\implies\cos y<0$$

$$1=|e^{iz}|=|e^{i(\cos y\pm i\sin y)}|=e^{\mp\sin y}\implies\sin y=0$$

Answer 2

$z=\dfrac{-b\pm\sqrt{b^2-4}}{2}$ and $|b|\leq 2$. So Re$z=-\dfrac{b}{2}$ and Im$z=\pm\dfrac{\sqrt{b^2-4}}{2}$.

Now, $$1>|e^z|=e^{Rez}\Rightarrow 0<b\leq 2$$ and $$1=|e^{iz}|=e^{-Imz}\Rightarrow b=\pm 2.$$

Answer 3

I've got it!

The issue couldn't have been any sillier.

I had forgotten the very specific meaning of the symbol $|$ $ $ $|$ for complex numbers..... Which of course means modulus of the complex number.

Usually, the modulus, when the complex number is written in the cartesian form $z=x+iy$, is $\sqrt{x^2+y^2}$, so it's simply the distance of the complex number from the origin.

However, when it's written in the exponential form $re^{iy}$, $ $ $r$ is then the modulus.

So, all the question was asking was to rearrange the inequality into the form $|re^{iy}|=r<1$, then deduce the value of $b$ from the value of $r$!

Ultimately, this gives $|e^{x+iy}|=|e^{x}e^{iy}|$, where $r=e^{x}$.
Hence, $|e^{x}e^{iy}|=e^{x}$, by simple definition of the operation $|$ $ $ $|$ for complex numbers.

So, stepping back to the initial inequality, we have $|e^z|=e^x<1$. Since $x=-b/2$, it follows that to satisfy (i), $b$ can be any number in the interval $(0, 2]$.

Part (ii) is quite easily deduced from the above.

PS:

Note to self -- never forget to revise topics you learnt months ago before starting to do exercises on them!