Modulus and Argument of cos(i) and -3i

In summary, the modulus for cos(i) is 1 and the argument is 0, while the modulus for -3i is 3 and the argument is 3π/2. The expression ½(e-1 + e) is in the form reix, with a modulus of 1 and an argument of 0.
  • #1
struggles
50
0

Homework Statement


Find the modulus and argument of
1) cos(i)
2) -3i

Homework Equations

The Attempt at a Solution


1) So for question 1 i tried
cos(i) = ½(ei2 + e-i2) = ½(e-1 + e) . However this doesn't (as far as I can see!) lead me to the right answer. I was aiming to get it in the form reix where r is the modulus and x the argument.

for 2) I basically just drew an argand diagram and found the radius to be 3 and the argument 3/4pi.

if you could point me in the right direction with 1 especially that would be greatly appreciated!
 
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  • #2
struggles said:

Homework Statement


Find the modulus and argument of
1) cos(i)
2) -3i

Homework Equations

The Attempt at a Solution


1) So for question 1 i tried
cos(i) = ½(ei2 + e-i2) = ½(e-1 + e) . However this doesn't (as far as I can see!) lead me to the right answer. I was aiming to get it in the form reix where r is the modulus and x the argument.

for 2) I basically just drew an argand diagram and found the radius to be 3 and the argument 3/4pi.

if you could point me in the right direction with 1 especially that would be greatly appreciated!
Concerning 1: what you have done is the right direction.
cos(i) = ½(e-1 + e) is correct.
Now the question is: why do you think this is not in the form reix?
 
  • #3
struggles said:

Homework Statement


Find the modulus and argument of
1) cos(i)
2) -3i

Homework Equations

The Attempt at a Solution


1) So for question 1 i tried
cos(i) = ½(ei2 + e-i2) = ½(e-1 + e) . However this doesn't (as far as I can see!) lead me to the right answer. I was aiming to get it in the form reix where r is the modulus and x the argument.

for 2) I basically just drew an argand diagram and found the radius to be 3 and the argument 3/4pi.
Why 3π/4? How many radians does it take to go around a circle one time?
if you could point me in the right direction with 1 especially that would be greatly appreciated!
 
  • #4
Samy_A said:
Concerning 1: what you have done is the right direction.
cos(i) = ½(e-1 + e) is correct.
Now the question is: why do you think this is not in the form reix?

I was thinking it contains no term with i. But would ½(e-1 + e) be equal to the modulus and have an argument of o?
 
  • #5
struggles said:
I was thinking it contains no term with i. But would ½(e-1 + e) be equal to the modulus and have an argument of o?
Exactly.
 
  • #6
SteamKing said:
Why 3π/4? How many radians does it take to go around a circle one time?
Oops 3π/2?
 
  • #7
Samy_A said:
Exactly.
Thank you!
 

FAQ: Modulus and Argument of cos(i) and -3i

What is the modulus of cos(i)?

The modulus of cos(i) is equal to 1. This can be derived using the Pythagorean identity, cos^2(x) + sin^2(x) = 1, where x is a complex number. Since cos(i) = cos(0) = 1, the modulus is also equal to 1.

What is the argument of cos(i)?

The argument of cos(i) is equal to pi/2 radians or 90 degrees. This can be derived by using the trigonometric identity, cos(x) = cos(-x). Since i is a purely imaginary number, -i has the same magnitude but opposite sign. Therefore, cos(i) = cos(-i) = cos(-pi/2) = cos(pi/2) = 0.

What is the modulus of -3i?

The modulus of -3i is equal to 3. This can be calculated using the absolute value function, which gives the distance of a complex number from the origin on the complex plane. Since -3i is located on the imaginary axis, its distance from the origin is 3 units.

What is the argument of -3i?

The argument of -3i is equal to -pi/2 radians or -90 degrees. This can be derived using the same logic as the argument of cos(i), where -3i can be written as 3i with a negative sign. Therefore, the argument of -3i is equal to the negative of the argument of 3i, which is pi/2.

Can the modulus and argument of a complex number be negative?

No, the modulus and argument of a complex number cannot be negative. The modulus is always a positive real number, while the argument is typically within the range of -pi to pi radians or -180 to 180 degrees. However, it is possible for the argument to be negative if the complex number is located in the lower half of the complex plane, where angles are measured clockwise instead of counterclockwise.

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