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Complex integer

  • Thread starter emutudeng
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  • #1
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z=6*e2,5i

Can anyone explain me ? The imaginary part = 3,59 and real part = -4,81

I tried e(x) = cos x + i sin x, but it does not help me.
 

Answers and Replies

  • #2
CAF123
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The complex number [itex] z [/itex] can be expressed in the form [itex] z = r(\cos(θ) + i\sin(\theta)), [/itex] where [itex] z = re^{i\theta}. [/itex] As long as theta is in radians, you should be able to read off the real and imaginary parts.
 
  • #3
Ray Vickson
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If you mean eix = cos(x) + i*sin(x), then that certainly gives you the correct answer; here you need to use x = 2.5 (N. American style), or x = 2,5 (Euro style).

Note, however, that the given answers are incorrect as exact statements; they are only approximations to the true values, which are approximately
real part ≈ -4.806861693281602289001016742804109986572
and
im part ≈ 3.590832864623738964311128213116973630222
to 40-digit accuracy. No matter how many digits we use we will never be able to write down the exact value.

RGV
 
  • #4
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How do I calcuate the values if x=2,5 ?
 
  • #5
Ray Vickson
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How do I calcuate the values if x=2,5 ?
What is stopping you from calculating cos(2,5) and sin(2,5)? (Remember, though, that the '2,5' is in radians, not degrees.)

RGV
 
  • #6
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okey tnx, i undrestood, but i have one other question that if there is no number in the middle for example z=e-5i then r = 1 ?
 
  • #7
CAF123
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okey tnx, i undrestood, but i have one other question that if there is no number in the middle for example z=e-5i then r = 1 ?
Yes.
 

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