Trig chart for the complex plane

i was trying to make a trig chart for the complex plane and accindently found 0>1. at pi/4, the hypotenuse is 0, but the real leg is 1. how is that? imaginary trig must be fun.

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
 

James R

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What's complex about trig functions of [itex]\theta = \pi/4[/itex]? It is not clear to me what you're doing. Hypotenuse of what? Real leg of what?
 
the legs are of the triangle. i thought that was implied, sorry.
 

James R

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So, are you talking about points on the unit circle in the complex plane?

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?
 
i drew the triangle and got it to be undefined. [tex]\frac{1}{\sqrt{2}}[/tex] is what i get on a real plane, but not on a complex one.
 

James R

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Please explain exactly what you are doing in your calculation.
 
i drew the triangle like i would on a real plane and found the trig funtions for pi/4
 

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?
 
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i don't have a scanner. i didn't do anything differently. i just used a complex plane instead of a real one.
 

James R

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If you didn't do anything differently, then there's no way you could get a different answer.

I guess we're done here, then.
 
yes there is...complex plane is not the real plane...
 

James R

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I'm sorry, but you don't seem to be able to explain yourself clearly enough for me to understand you. Maybe somebody else will come along who will know what you're talking about.

Good luck.
 
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?
 
Last edited:
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?
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.
 
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no your answer isn't correct. unless your understanding of complex or perhap hypoteneuse is muttle you can't get an answer of 0. James already stated the modulus is 1...the lenght of boths sides of your right angle are invsqrt(2)-> sqr then add you get 1...so your lenght is 1 in complex OR real...

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)
 

Integral

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Since the complex plane is isomorphic to R2, I share James confusion about what you are doing. How did you come up with a zero length hypotenuse? IIRC, the hypotenuse is always 1 when working in a unit circle. It does not matter if you do it in the complex plane or R2
 

arildno

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yourdaddyspoggystick:

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?
 

matt grime

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So, what he appears to be doing, worknig in the first uadrant:

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.
 

arildno

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Matt Grime's suggestion seems to be the most likely explanation of the pogostick's mistake.
 
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yourdadonapogostick said:
i was trying to make a trig chart for the complex plane and accindently found 0>1. at pi/4, the hypotenuse is 0, but the real leg is 1. how is that? imaginary trig must be fun.

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
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.

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]
 
The kid was probably like "ok, on the complex plane, let's say you go up a distance of i and right a distance of 1." Then he took the "modulus" and since i^2=-1, he got a "modulus" of zero.

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.
 

SGT

I think I understand your error. You calculated the real side as [tex]\frac{\sqrt{2}}{2}[/tex] and the imaginary side as [tex]\frac{\sqrt{2}i}{2}[/tex]. So, the real side squared plus the imaginary side squared sum to zero.
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.
 
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Here lies the mistake, hermitian instead of euclidean
 

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