Log of negative

which is that e^y on its own represents a complex number of unit modulus

Why do you think that?

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

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

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?