Complex Numbers - a few questions.

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I have a few questions about complex numbers so I figured I may as well put them all into one big thread

1. When doing calculus with complex numbers, is i just treated as another constant? I haven't seen much mention of complex calculus anywhere, so I figured I should ask For example, what is \dfrac{d}{dx} e^{-3ix}

2. Can you extend the natural logarithm to complex and negative numbers by expressing in exponential form and using log laws? For example  \text{ln} (-5) = \text{ln} \left( 5 e^{ i \pi} \right) = \text{ln} 5 + \text{ln} \left( e^{ i \pi} \right)

3. Are equations with complex coefficients solved in the same way as equations with real coefficients? For example, if you had (2+3i)x^2+(4+3i)x+(1-2i), would we just solve this using the quadratic formula?

Thanks for helping!

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Smaug123
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(Original post by majmuh24)
I have a few questions about complex numbers so I figured I may as well put them all into one big thread

1. When doing calculus with complex numbers, is i just treated as another constant? I haven't seen much mention of complex calculus anywhere, so I figured I should ask For example, what is \dfrac{d}{dx} e^{-3ix}

2. Can you extend the natural logarithm to complex and negative numbers by expressing in exponential form and using log laws? For example  \text{ln} (-5) = \text{ln} \left( 5 e^{ i \pi} \right) = \text{ln} 5 + \text{ln} \left( e^{ i \pi} \right)

3. Are equations with complex coefficients solved in the same way as equations with real coefficients? For example, if you had (2+3i)x^2+(4+3i)x+(1-2i), would we just solve this using the quadratic formula?

Thanks for helping!
1. Yes. \dfrac{d}{dx} 3^{-3i x} = -3i e^{-3i x}. This is because the chain rule can be proved in the complex numbers (in almost exactly same way as in the reals).

2. Yes, sort of. \log(z) = \text{ln}(|z|) + i \arg(z). However, you have to be careful because you no longer get the property that \log(ab) = \log(a)+\log(b): take a = b = e^{\frac{3}{4} i \pi}. Because the log requires a consistent definition of the argument, and because the argument doesn't necessarily behave nicely on multiplication, we therefore have that log doesn't necessarily behave nicely on multiplication.

3. Yes, exactly. We can factorise equations in the same way over the complex numbers as we can over the reals.
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#3
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To add to the above:

1. Your example is "calculus with complex numbers" and not "calculus in the complex numbers". The former is not terribly interesting, whereas the latter [i.e. complex analysis] is fascinating and powerful. Most of the neatness behind complex analysis stems from the notion of differentiability: although the definition is essentially the same as in the real case, it is in fact a far stronger requirement [essentially because we need the limit to exist and agree when approached from all directions, and not just two]. For instance, if f:\;\mathbb{C}\to\mathbb{C} is once-differentiable (holomorphic) it is automatically infinitely differentiable. Or, if f is analytic in a simply connected domain \mathcal{D} then its integral along any contour in \mathcal{D} is automatically 0. There are no parallels to these in real analysis.

2. \log z is multivalued because \arg z is multivalued. The remedy is to choose a branch (by making a branch cut i.e. making a slice in the complex plane so as to restrict a multivalued function to a single-valued one). The traditional choice for \log is to restrict the argument to (-\pi,\pi] in which case we write \text{Log}\,z=\log |z|+i\text{Arg}\,z
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