- #1

yourdadonapogostick

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at [tex]\theta=\frac{\pi}{4}[/tex], i got:

sin=undefined

cos=undefined

tan=i

csc=0

sec=0

cot=-i

it is in the first quadrant and cot is negative...something is wonky

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- Thread starter yourdadonapogostick
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- #1

yourdadonapogostick

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at [tex]\theta=\frac{\pi}{4}[/tex], i got:

sin=undefined

cos=undefined

tan=i

csc=0

sec=0

cot=-i

it is in the first quadrant and cot is negative...something is wonky

- #2

James R

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- #3

yourdadonapogostick

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the legs are of the triangle. i thought that was implied, sorry.

- #4

James R

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By definition, all points on that circle have modulus 1. The real part of any point on the circle is given by [itex]\cos \theta[/itex], where [itex]\theta[/itex] is the angle between the real axis and the line joining the origin to the point on the circle.

[itex]\cos \pi/4 = 1/\sqrt{2}[/itex]

Is this what you're talking about, or something else?

- #5

yourdadonapogostick

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- #6

James R

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Please explain exactly what you are doing in your calculation.

- #7

yourdadonapogostick

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i drew the triangle like i would on a real plane and found the trig funtions for pi/4

- #8

James R

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So, what did you do differently on the complex plane compared to the real (x,y) plane?

The value of [itex]\cos \pi/4[/itex] never changes.

Also, you talk about having a hypotenuse of zero with a real leg of 1. Are you talking about a right-angled triangle? Because any triangle with a non-zero side adjacent to the right angle will have a non-zero hypotenuse - otherwise it isn't a right angled triangle.

Can you post your diagram?

The value of [itex]\cos \pi/4[/itex] never changes.

Also, you talk about having a hypotenuse of zero with a real leg of 1. Are you talking about a right-angled triangle? Because any triangle with a non-zero side adjacent to the right angle will have a non-zero hypotenuse - otherwise it isn't a right angled triangle.

Can you post your diagram?

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- #9

yourdadonapogostick

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- #10

James R

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I guess we're done here, then.

- #11

yourdadonapogostick

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yes there is...complex plane is not the real plane...

- #12

James R

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Good luck.

- #13

yourdadonapogostick

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i don't understand what you don't understand.

i drew a triangle in the first quadrant with a central angle being pi/4. when i found the lenth of the hypotenuse, it was 0. what is there not to understand?

i drew a triangle in the first quadrant with a central angle being pi/4. when i found the lenth of the hypotenuse, it was 0. what is there not to understand?

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- #14

Stephan Hoyer

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How did you obtain the value "0" then? Clearly the hypotenuse as shown in your drawing was of non-zero length, so you did some sort of calculation to obtain that value.yourdadonapogostick said:i don't understand what you don't understand.

i drew a triangle in the first quadrant with a central angle being pi/4. when i found the lenth of the hypotenuse, it was 0. what is there not to understand?

- #15

neurocomp2003

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How on Earth did you get the hyp =0?

REAL: invsqrt(2)*i +invsqrt(2)*j

CMPLX: invsqrt(2)*I +invsqrt(2)*R

the coordinates are identical, thus the lengths of the sides are the same length.

lenght of a complex vector is its modulus. sqrt(a^2+b^2)

- #16

Integral

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- #17

arildno

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Remember that the squared modulus of a complex number z is given by zz', where z' is the complex conjugate of z. It is not given by zz; perhaps it was here that you erred?

- #18

matt grime

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take the points (0,0) (x,0) (x,yi) where x^2+y^2=1 and take the triangle these define.

he now chooses to define the trig functions as the ratios such as yi/x etc and believes that the pythagorus allows us to deduce that x^2+(yi)^2 is the length of the hypoteneuse. since those are not the definitino of complex trig funcions, or the "length of a hypoteneuse" we can see the mistake.

- #19

arildno

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Matt Grime's suggestion seems to be the most likely explanation of the pogostick's mistake.

- #20

chronon

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Given a right angled triangle with sides a and b, the angle is given by arctan(b/a), and then the sine and cosine can be calculated. This still makes sense if a and b are complex numbers. However there seems to be a problem with a=1 b=i, since I think arctan(z) in singular at z=i.yourdadonapogostick said:

at [tex]\theta=\frac{\pi}{4}[/tex], i got:

sin=undefined

cos=undefined

tan=i

csc=0

sec=0

cot=-i

it is in the first quadrant and cot is negative...something is wonky

Of course for a complex number z, the angle arg(z) is totally different from the 'angle' arctan(z) (which will be a complex number). Also arg(i) =[tex] \pi/2[/tex] not [tex] \pi/4[/tex]

- #21

philosophking

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But sqrt(i^2) here is not the _length of i_. sqrt( |i^2| ) is. So the modulus would be sqrt ( 1^2 + |i^2| ) = sqrt(2) and everything works out fine.

- #22

But the square of the hypotenuse is the sum of the squares of the moduli of the sides. The modulus of the imaginary side is [tex]\frac{\sqrt{2}}{2}[/tex], so the length of your hypotenuse is 1, as everybody has told you.

- #23

Maxos

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Here lies the mistake, hermitian instead of euclidean

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