Simplifying Complex Arithmetic: Converting cos(1+i) to Cartesian and Euler Forms

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In summary, the equation for cosine and sine in Euler form is cos(i)=-sin(i)sinh(i). The hyperbolic functions are related to the polar form of these functions in the same way that the polar form is related to the Cartesian form.
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Homework Statement



I have a very simple complex arithmetic question.

How do I express the quantity cos(1+i) in Cartesian (a+ib) and Euler(re^i*theta)

Is this the right track?:

[tex]cos(1+i)={e^{i(1+i)}\over2}+{e^{-i(1+i)}\over2}[/tex]

I know that:

[tex]cos(1+i)=cos(1)cos(i)-sin(1)sin(i)[/tex]

but I am not sure how to get the "i's" out of the cosine and sine.
 
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  • #2
Well, sure you are on the right track. The same formula you used before. cos(i)=(e^(i*i)+e^(i*(-i))/2=(e^(-1)+e^1)/2. That's cosh(1). Use similar expression for sin(i).
 
  • #3
Dick said:
Well, sure you are on the right track. The same formula you used before. cos(i)=(e^(i*i)+e^(i*(-i))/2=(e^(-1)+e^1)/2. That's cosh(1). Use similar expression for sin(i).


So then as far as cartesian form goes:


[tex]cos(1+i)=cos(1)cosh(1)-i(sin(1)sinh(1))[/tex]

If that is correct, what is the relationship between the hyperbolic functions and the notation:

[tex]re^{i\theta}[/tex]
 
  • #4
Now you've got cos(1+i) written in the form a+ib where a and b are real. Now it's the usual relation with the polar form. r=sqrt(a^2+b^2), theta=arctan(b/a), right? There's nothing terribly spooky or mysterious going on here.
 
  • #5
Thanks Dick. I don't know why this stuff cooks my noodle so much. I feel as though I cannot wrap my head around it. I always think there is some Identity that I am not taking into account or that my answers are not nearly simplified enough.

One last question. Would you consider it correct to write the equation in Euler form this way:


[tex]\sqrt{(cos(1)cosh(1))^2+(sin(1)sinh(1))^2}e^{i(arctan{(sin(1)sinh(1)\over(cos(1)cosh(1)})}[/tex]
 
  • #6
That's about as good as I could come up with. But you are missing a sign in the argument. You could also write tan(1)*tanh(1) in the argument. But that's not all that much of a real simplification, is it?
 

FAQ: Simplifying Complex Arithmetic: Converting cos(1+i) to Cartesian and Euler Forms

What is the difference between Cartesian and Euler forms in complex arithmetic?

Cartesian form represents complex numbers in terms of their real and imaginary components, while Euler form represents them in terms of their magnitude and phase angle.

How do you convert a complex number from Cartesian to Euler form?

To convert from Cartesian form (a + bi) to Euler form (re), use the formulas r = √(a2 + b2) and θ = tan-1(b/a). The magnitude (r) is the distance from the origin to the point on the complex plane, and the phase angle (θ) is the angle formed by the complex number and the positive real axis.

Can you convert a complex number from Euler to Cartesian form?

Yes, you can convert from Euler form (re) to Cartesian form (a + bi) using the formulas a = r cosθ and b = r sinθ. Note that r is the magnitude and θ is the phase angle of the complex number.

What is the significance of using Euler form in complex arithmetic?

Using Euler form can simplify complex arithmetic by allowing us to use trigonometric identities and operations, making calculations more efficient and easier to understand.

How do you convert a complex number from polar form to rectangular form?

To convert from polar form (rθ) to rectangular form (a + bi), use the formulas a = r cosθ and b = r sinθ. Note that r is the magnitude and θ is the angle in radians.

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