Solve Imaginary Numbers: 2*EXP(i*pi/3) -> 1+sqt(3)i

[VeganGirl]

Homework Statement

Write 2*EXP(i*pi/3) in the form $$\alpha$$ + i$$\beta$$
Answer is given = 1 + sqt(3)i

Homework Equations

The Attempt at a Solution

I'm supposed to turn this exponential form of imaginary number into a standard form in order to solve an ODE.

I have no idea how they got 1+sqt(3)i from the exponential form of the complex number given. I mean, where did the exponential go?

There must be a simple way of converting, but I just can't seem to find it in the text or online. Please help!
And thank you in advance!

 

[Pengwuino]

Do you know what Euler's equation is?

 

[HallsofIvy]

$$e^{i\theta}= cos(\theta)+ i sin(\theta)$$.

$z= re^{i\theta}= r(cos(\theta)+ i sin(\theta)$.

And, if z= x+ iy, then \(\displaystyle r= \sqrt{x^2+ y^2}\) and \(\displaystyle tan(\theta)= \frac{y}{x}\) as long as x is not 0.

If you are working with "exponential form", I would be very surprised if you did not already know that.

 

[VeganGirl]

Do you know what Euler's equation is?

Firstly, thanks for your reply.
And yes... Eular's equation says...
e^(i*beta*t) = cos (beta*t) + i*sin (beta*t)
But I wasn't sure how to use this equation since t is missing in the question...

 

[JonF]

But I wasn't sure how to use this equation since t is missing in the question...

Write down Euler’s Equation and replace all of the t’s with pi/3

 

[VeganGirl]

$$e^{i\theta}= cos(\theta)+ i sin(\theta)$$.

$z= re^{i\theta}= r(cos(\theta)+ i sin(\theta)$.

And, if z= x+ iy, then \(\displaystyle r= \sqrt{x^2+ y^2}\) and \(\displaystyle tan(\theta)= \frac{y}{x}\) as long as x is not 0.

If you are working with "exponential form", I would be very surprised if you did not already know that.

Thank you for your reply.
I'm not a math genius (as you could probably tell) and imaginary numbers, especially tend to throw me off.

But your equations help a lot.
Frankly, I've never seen this equation. z = re^(i*theta) = r (cos (theta) + i sin (theta).

Anyway, thanks again for your help

 

[JonF]

That’s the polar form of complex numbers, as hallsofivy said it would be very odd if they gave you exponential form but not polar.

 

[VeganGirl]

Write down Euler’s Equation and replace all of the t’s with pi/3

Hi. Thanks for your reply.

I'm doing another question like this and this one has a variable "t" in it.

Q) 2e^(i*$$\sqrt{2}$$*t) => Write in Standard Form

Using Euler's Equation, I have

-> 2*(cos $$\sqrt{2}$$*t + i sin $$\sqrt{2}$$ t)

How do I simplify this? Since the angle $$\sqrt{2}$$ is not in the unit circle?

 

[JonF]

Standard form in your book is probably defined as z = x + yi.

Where x and y are reals.

So z = 2cos(2^1/2 * t) + 2sin(2^1/2 * t)*i would be standard form, but how you wrote it is typically accepted also.

 

[Inferior89]

Hi. Thanks for your reply.

I'm doing another question like this and this one has a variable "t" in it.

Q) 2e^(i*$$\sqrt{2}$$*t) => Write in Standard Form

Using Euler's Equation, I have

-> 2*(cos $$\sqrt{2}$$*t + i sin $$\sqrt{2}$$ t)

How do I simplify this? Since the angle $$\sqrt{2}$$ is not in the unit circle?

Wait.. wait. Did you say the angle sqrt(2) is not in the unit circle? The unit circle has angles between 0 and 2 pi which is about 0 to 6.2. sqrt(2) is about 1.4.

 

[HallsofIvy]

Standard form in your book is probably defined as z = x + yi.

Where x and y are reals.

So z = 2cos(2^1/2 * t) + 2sin(2^1/2 * t)*i would be standard form, but how you wrote it is typically accepted also.

Don't even think that! A major reason for defining sine and cosine in terms of the "unit circle", rather than in terms of right triangles, is that we can keep going around the circle as many times as we want or we can go counter-clockwise rather than clockwise making the argument negative.

sin(x) and cos(x) are defined for all real numbers x (and can be extended to complex numbers).

For this particular problem, $2(cos(t\sqrt{2})+ i sin(t\sqrt{2}))$ is the "standard form". You don't need to simplify it.

 

[VeganGirl]

Thanks everyone! You guys have been a great help!