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Big-Daddy
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I am trying to use the exponential form of complex numbers to evaluate ln(-9) or in general ln(a+bi). Taking ln(-9) as my example, I set y=ln(-9) so -9 = e^y. Now this resembles the form I am used to in some ways, but it still brings in its own problem, which is that e^y on its own represents a complex number of unit modulus, whereas -9 has modulus = 9. So we really would prefer to say -9 = 9 * e^y where y=i*theta, and then evaluate theta to find the complex solution, but we can't just stick 9 on the RHS from nowhere surely. So what do I do?
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Notnek
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(Original post by Big-Daddy)
which is that e^y on its own represents a complex number of unit modulus
Why do you think that?
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Big-Daddy
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(Original post by notnek)
Why do you think that?
because I want to find the complex number solution and as far as I know, in the complex number solution of exponential form r*e^(i*theta), r represents modulus (magnitude) of the complex number - and here r=1 if we just have e^(i*theta) on the RHS
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davros
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(Original post by Big-Daddy)
I am trying to use the exponential form of complex numbers to evaluate ln(-9) or in general ln(a+bi). Taking ln(-9) as my example, I set y=ln(-9) so -9 = e^y. Now this resembles the form I am used to in some ways, but it still brings in its own problem, which is that e^y on its own represents a complex number of unit modulus, whereas -9 has modulus = 9. So we really would prefer to say -9 = 9 * e^y where y=i*theta, and then evaluate theta to find the complex solution, but we can't just stick 9 on the RHS from nowhere surely. So what do I do?
You seem to be making a bit of a meal of this.

You can write ln(-9) = ln(9) + ln(-1) - there is no difficulty in evaluating ln(9) so your only issue is in choosing the principal value of ln(-1) which hopefully you should recognize from Euler's famous equation
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Big-Daddy
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(Original post by davros)
You seem to be making a bit of a meal of this.

You can write ln(-9) = ln(9) + ln(-1) - there is no difficulty in evaluating ln(9) so your only issue is in choosing the principal value of ln(-1) which hopefully you should recognize from Euler's famous equation
Thanks, but this method seems a bit limited in its flexibility. How about ln(a + bi) more generally? I figure you'd need to use the complex method something like I indicated in the OP to get to the answer?
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james22
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(Original post by Big-Daddy)
because I want to find the complex number solution and as far as I know, in the complex number solution of exponential form r*e^(i*theta), r represents modulus (magnitude) of the complex number - and here r=1 if we just have e^(i*theta) on the RHS
This is only true when theta is real. e^(a+bi)) has modulus e^a if a and b are real.
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davros
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(Original post by Big-Daddy)
Thanks, but this method seems a bit limited in its flexibility. How about ln(a + bi) more generally? I figure you'd need to use the complex method something like I indicated in the OP to get to the answer?
Same principle!

a + bi = re^{i\theta} = r(\cos \theta + i\sin \theta)

Now solve for r and theta (it's the same sort of idea as when you put a trig expression into harmonic form Rsin(x+a))

Then ln(a +bi) = ln r + i\theta
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