complex numbers
I'm stuck on a question in relation with complex numbers:
If $z\neq0$, and that
$$\left\vert{\frac{z+1}{z-1}}\right\vert=1,$$
prove that $z$ is purely imaginary.
I tried breaking the modulus up into two separate parts, and then multiplying both sides. Then I squared both sides and used the formula $\vert{z}\vert^2=z\bar{z}$, and replaced $z$ with $a+bi$. But it seems that no matter what manipulation I do, it either turns real or doesn't make sense.
Any help is much appreciated!! Thanks!
To briefly address your specific question about "duality" first, there is no "duality" (not sure what is precisely meant by that term here) between complex exponentiation and complex multiplication, unless the arguments of both are real. This is because complex multiplication is commutative; complex exponentiation is not - neither is real exponentiation $(e^{ix}$ is not the same as $ix^{e}$). Take the second line:
$$c^{a+bi}\cdot z \leftrightarrow (a+bi)\cdot z $$ $$ \text{stretching and rotating} \leftrightarrow \text{stretching and rotating} $$
Yes, $(a+bi)\cdot z$ corresponds to a transformation consisting of some stretching and some rotation, but the stretching is due to both multiplying by $a$ and multiplying by $bi$.
Now, let's break down intuitively what different operations on the complex plane are (I know you said you already went over some of this in your question, but I think it will lead into the explanation for complex exponentiation nicely).
Complex addition is the same as vector addition: we add the components. Think of this intuitively by imagining each complex number as an arrow: adding two complex numbers is like sticking one of their arrows on the end of the other. Another way: think of adding a complex number not as a static operation, but as a transformation. Adding the number $(a+bi)$ is the same as shifting the origin of the complex plane onto the point $(-a-bi)$. Take a second to imagine why that's true. Thus $(a+bi)$ is a function with respect to addition, which maps every point in the complex plane to another point in the complex plane, (a+bi) away.
Complex multiplication corresponds to both a stretch and a rotation (usually). $$(a+bi)\cdot(c+di)=(ac-bd)+(ad+bc)$$
A better way to think about this is in terms of Euler's formula: represent your two complex numbers as polar coordinates, and multiplication becomes much clearer: $$r_1 e^{i\theta_1}\cdot r_2 e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}$$
So we can imagine complex multiplication of two numbers is taking the angles they make with the x axis, adding those two angles to get the angle of your new number, then multiplying the magnitudes of the two original numbers to get the magnitude of your new number. Think of complex multiplication by $(a+bi)$ as two, very dynamic transformations composted together: first stretching the entire complex plane by a factor of $\sqrt{a^2+b^2}$, then rotating the entire complex plane by a factor of $\tan^{-1}(\frac{b}{a})$. Thus $(a+bi)$ can also be thought of as a function with respect to multiplication: it maps every point in the complex plane to another point in the complex plane by combination of a rotation and a stretch.
What kind of function is "complex exponentiation"? We define it as follows: $a^{b+c\cdot i}$=$a^{b}\cdot a^{c\cdot i}$ where $a^{c \cdot i} = e^{c\cdot i \log a}$, etc. based on Euler's formula. $e^{c\cdot i \log a}$ is $e$ raised to a complex number and is thus a rotation. Note that we can imagine complex exponentiation again as a sort of dynamic transformation, squishing and mapping the complex plane to a new location.
Let's break this down into two cases. The first is the only one your question asks about.
1) The base of the exponent is real. As can be seen from above, every "complex exponentiation" is a transformation of the complex plane consisting of two transformations: first, a stretch by some factor, and then a rotation. This is very similar to complex multiplication, which begs the question, when do these two things behave in the same way? You called it "duality," I'm not going to call it that because that word means something specific in linear algebra, I'll just call this similarity "similarity."
Multiplying $x$ by $(a+bi)\rightarrow$ Stretch by $\sqrt{a^2+b^2}$, rotate by $\tan^{-1}(\frac{b}{a})$ radians.
Raising $x$ to the $(a+bi)$ power $\rightarrow$ Stretch by $x^a$, rotate by $b\cdot i \log x$ radians.
Note that the two are very dissimilar - exponentiation is dependent on the base whereas multiplication is not. This is a result of the effect that multiplication by a complex number is something called a linear transformation (https://en.wikipedia.org/wiki/Linear_map), raising something to a complex power is most certainly not.
2) The base of the exponent is complex. This gets a little more complicated, because raising a complex number to a real power corresponds in part to a rotation, so separating which parts are the rotation and which parts are the stretching is a little annoying and won't give much insight here. Complex exponentiation of complex numbers is really funky, giving rise to all sorts of weird fractal shapes when we consider which complex numbers get really large when we raise them to a complex power, and which ones don't - this is related to how the Mandelbrot set is formed (https://en.wikipedia.org/wiki/Mandelbrot_set). The point is that, again, the magnitude of the stretch and rotation is dependent on x, meaning that this is not a linear transformation.
If you want to gain an intuition for how complex exponentiation of complex numbers works, I recommend you play around with some functions with this grapher: http://davidbau.com/conformal/#z
So, to return to what I mentioned at the beginning, the reason why you're not seeing a "duality" in case 3 is that there's no "duality" in case 2 to begin with - yes, both complex exponentiation and complex multiplication correspond to both a rotation and a stretch, but the rotation and stretch for each behave in very different ways. Complex multiplication is a linear transform, complex exponentiation is not.
I also enjoy brilliant.org's courses; if you're interested, I would recommend you check out their course on linear algebra next (https://brilliant.org/courses/linear-algebra/). This is the first answer I've actually posted. I would love feedback from anyone if they have it.
$y$ must be a real number.
Since there is no real number satisfying $y^4-4y^2+7=0$ and $y(3-y^2)=0$ at the same time, there is no purely imaginary root of $z^4 + z^3 + 4z^2+3z+7 = 0$.
Best Answer
If $|z-1|=|z+1|$, then writing $z=x+iy$ and squaring both sides we get $$ (x-1)^2+y^2=(x+1)^2+y^2$$ which implies that $x=0$. If $z\neq 0$, this means that $z$ is purely imaginary.
There is also a nice geometric interpretation: $|z-1|$ is the distance from $z$ to $1$, and $|z+1|$ is the distance to $-1$. If these are equal, then $z$ must lie on the perpendicular bisector of the line segment connecting $1$ and $-1$, which is the line $x=0$.