MHB How to Prove a Complex Number Equation and Its Trajectory Forms a Circle?

[WMDhamnekar]

My attempt:
Let us put $\frac{1}{i+t} = \frac{1+e^{is}}{2i} \Rightarrow \frac{2i}{i+t} -1= e^{is}$

So, $\cos{s}- i\sin{s}= \frac{2i}{i+t} - 1,\Rightarrow \cos^2{(s)} - \sin^2{(s)} = \frac{-2}{(i+t)^2} +1 -\frac{4i}{i+t}$

After doing some more mathematical computations, I got $\cos{s}= \frac{t}{i+t}$ Now how to answer this question? i-e how to prove $\frac{1}{i+t}= \frac{1+e^{is}}{2i}$ and the trajectory for arbitrary $\alpha, \beta \in \mathbb{C} $ forms a circle?

 

[I like Serena]

My attempt:
Let us put $\frac{1}{i+t} = \frac{1+e^{is}}{2i} \Rightarrow \frac{2i}{i+t} -1= e^{is}$

The expression means that we have an imaginary number on the unit circle.
That is, it has magnitude 1 and can have any angle.

So if we can prove that $\left|\frac{2i}{i+t} -1\right|\overset ?= 1$, we're basically done.

 

[soloenergy]

Firstly, to prove $\frac{1}{i+t}= \frac{1+e^{is}}{2i}$, we can use the fact that $e^{is} = \cos{s} + i\sin{s}$. Substituting this into the equation, we get:

$\frac{1}{i+t} = \frac{1+\cos{s} + i\sin{s}}{2i}$

Using the fact that $\cos^2{s} + \sin^2{s} = 1$, we can simplify the right side to:

$\frac{1}{i+t} = \frac{1+\cos{s}}{2i} + \frac{i\sin{s}}{2i}$

Simplifying further, we get:

$\frac{1}{i+t} = \frac{1}{2i} + \frac{i\sin{s}}{2i}$

And finally, using the fact that $i^2 = -1$, we get:

$\frac{1}{i+t} = \frac{1}{2i} + \frac{i}{2}$

Which can be further simplified to:

$\frac{1}{i+t} = \frac{1+e^{is}}{2i}$

Therefore, we have proven that $\frac{1}{i+t}= \frac{1+e^{is}}{2i}$.

Next, to prove that the trajectory for arbitrary $\alpha, \beta \in \mathbb{C}$ forms a circle, we can use the fact that the equation for a circle is $x^2 + y^2 = r^2$, where $r$ is the radius of the circle.

In this case, we have $\alpha = x$ and $\beta = y$. Substituting these values into the equation, we get:

$x^2 + y^2 = r^2$

Now, using the equation we proved earlier, we can substitute $\frac{1}{i+t}$ for $x$ and $\frac{1+e^{is}}{2i}$ for $y$. This gives us:

$(\frac{1}{i+t})^2 + (\frac{1+e^{is}}{2i})^2 = r^2$

After simplifying, we get:

$\frac{1}{(i+t)^2} + \frac{1+2e^{