Complex number question

• Dec 2, 2021

• #1

WMDhamnekar

MHB

376

28

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?

 

Physics news on Phys.org

• Research suggests how turbulence can be used to generate patterns
• High-quality microwave signals generated from tiny photonic chip
• Quantum talk with magnetic disks

• Dec 2, 2021

• #2

I like Serena

Homework Helper

MHB

16,336

258

Dhamnekar Winod said:

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.

 

Last edited: Dec 2, 2021

• Dec 2, 2022

• #3

soloenergy

64

0

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^{

 

1. What is a complex number?

A complex number is a number that contains both a real part and an imaginary part. It is written in the form a + bi, where a is the real part and bi is the imaginary part, with i being the imaginary unit.

2. How do you add or subtract complex numbers?

To add or subtract complex numbers, you simply add or subtract the real parts and the imaginary parts separately. For example, (3 + 2i) + (5 + 4i) = (3 + 5) + (2i + 4i) = 8 + 6i.

3. What is the difference between a real number and a complex number?

A real number is any number that can be represented on a number line, including both positive and negative numbers. A complex number, on the other hand, includes both a real part and an imaginary part, and cannot be represented on a number line.

4. How do you multiply complex numbers?

To multiply complex numbers, you use the FOIL method (First, Outer, Inner, Last). For example, (3 + 2i)(5 + 4i) = 3(5) + 3(4i) + 2i(5) + 2i(4i) = 15 + 12i + 10i + 8i² = 15 + 22i - 8 = 7 + 22i.

5. Can complex numbers be divided?

Yes, complex numbers can be divided. To divide complex numbers, you use the conjugate of the denominator to rationalize the fraction. For example, (3 + 2i)/(5 + 4i) = ((3 + 2i)(5 - 4i))/((5 + 4i)(5 - 4i)) = (15 + 10i - 12i + 8i²)/(25 - 16i²) = (7 - 2i)/41.