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Complex number question

Dhamnekar Winod

Active member
Nov 17, 2018

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?

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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.
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