Why we took the positive only?

[YehiaMedhat]

I have a question in complex analysis sheet to prove that [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath], but when trying to reach the solution my self, I get this [imath]\cos^{-1}(z) = -i\ln(z\pm\sqrt{z^2-1})[/imath].

Since the [imath]\cos^{-1}(z) = -i\ln(z-\sqrt{z^2-1})[/imath] won't yield a negative number, so it doesn't get rejected in the domain of the [imath]\ln[/imath] function, why the final answer just give the answer like this [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath]???

 

[blamocur]

I believe you are right. But it sounds like the sheet wants you to prove that one equality, not to find all expressions for [imath]\cos^{-1}(z)[/imath].

 

[mario99]

I have a question in complex analysis sheet to prove that [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath], but when trying to reach the solution my self, I get this [imath]\cos^{-1}(z) = -i\ln(z\pm\sqrt{z^2-1})[/imath].

Since the [imath]\cos^{-1}(z) = -i\ln(z-\sqrt{z^2-1})[/imath] won't yield a negative number, so it doesn't get rejected in the domain of the [imath]\ln[/imath] function, why the final answer just give the answer like this [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath]???

It is [imath]z\pm\sqrt{z^2-1}[/imath]. But by definition they just write [imath]z + \sqrt{z^2-1}[/imath].

When you simplify, you may want to write it again [imath]\pm[/imath]. I will give you an example.

Let [imath]z = \frac{5}{3}[/imath] and find [imath]\cos^{-1}\frac{5}{3}[/imath].

You will say by definition, [imath]\cos^{-1}(z) = -i\ln\left(z+\sqrt{z^2-1}\right)[/imath]

[imath]\cos^{-1}(\frac{5}{3}) = -i\ln\left(\frac{5}{3}+\sqrt{\frac{16}{9}}\right)[/imath]

But then, you will write:

[imath]\cos^{-1}(\frac{5}{3}) = -i\ln\left(\frac{5}{3}\pm\frac{4}{3}\right) \ \ \ \ \ \ [/imath] (If you want to find all solutions.)

---------------
For proofs, you can choose either one:
[imath]z+\sqrt{z^2-1}[/imath]
[imath]z-\sqrt{z^2-1}[/imath]
because both satisfy the equation.

 

[YehiaMedhat]

So it's just fine, and not a conspiracy in the text books
Thanks guys

 

[Dr.Peterson]

I have a question in complex analysis sheet to prove that [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath], but when trying to reach the solution my self, I get this [imath]\cos^{-1}(z) = -i\ln(z\pm\sqrt{z^2-1})[/imath].

Since the [imath]\cos^{-1}(z) = -i\ln(z-\sqrt{z^2-1})[/imath] won't yield a negative number, so it doesn't get rejected in the domain of the [imath]\ln[/imath] function, why the final answer just give the answer like this [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath]???

There are subtleties here that have only been hinted at.

First, your question in the title is wrong: the "positive" square root is not actually positive, since it is not, in general, a real number at ll! Complex numbers in general are neither positive nor negative.

Second, in the complex world, "the square root" is not quite what it seems to be; every non-zero number has two square roots, and even if you define a "principal" root, that is not always significant. I would either take [imath]\sqrt{z}[/imath] as implicitly including both of [imath]\pm\sqrt{z}[/imath], or always write the latter explicitly. In your context (course?), how has [imath]\sqrt{z}[/imath] been defined?

Third, when you mention the domain of the [imath]\ln[/imath] function "rejecting a negative number", are you assuming that the argument of [imath]\ln[/imath] in your expression is a real number??

Finally, is this asking only for the principal value of the inverse cosine?

I'd like to see the actual question as given to you, and the contextual answers to my questions here. (Is z here meant to be a real number? What is a positive root? What is the square root? What is the domain of [imath]\ln[/imath]? Does [imath]\cos^{-1}(z)[/imath] refer to the principal value?)

One final question: Have you considered how [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath] and [imath]\cos^{-1}(z) = -i\ln(z-\sqrt{z^2-1})[/imath] are related, by expressing one in terms of the other? (Hint: use conjugates.)

 

[YehiaMedhat]

There are subtleties here that have only been hinted at.

First, your question in the title is wrong: the "positive" square root is not actually positive, since it is not, in general, a real number at ll! Complex numbers in general are neither positive nor negative.

Second, in the complex world, "the square root" is not quite what it seems to be; every non-zero number has two square roots, and even if you define a "principal" root, that is not always significant. I would either take [imath]\sqrt{z}[/imath] as implicitly including both of [imath]\pm\sqrt{z}[/imath], or always write the latter explicitly. In your context (course?), how has [imath]\sqrt{z}[/imath] been defined?

Third, when you mention the domain of the [imath]\ln[/imath] function "rejecting a negative number", are you assuming that the argument of [imath]\ln[/imath] in your expression is a real number??

Finally, is this asking only for the principal value of the inverse cosine?

I'd like to see the actual question as given to you, and the contextual answers to my questions here. (Is z here meant to be a real number? What is a positive root? What is the square root? What is the domain of [imath]\ln[/imath]? Does [imath]\cos^{-1}(z)[/imath] refer to the principal value?)

One final question: Have you considered how [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath] and [imath]\cos^{-1}(z) = -i\ln(z-\sqrt{z^2-1})[/imath] are related, by expressing one in terms of the other? (Hint: use conjugates.)

I really appreciate it, but I feel unsafe mathematically

 

[Steven G]

I have a question in complex analysis sheet to prove that [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath], but when trying to reach the solution my self, I get this [imath]\cos^{-1}(z) = -i\ln(z\pm\sqrt{z^2-1})[/imath].

Since the [imath]\cos^{-1}(z) = -i\ln(z-\sqrt{z^2-1})[/imath] won't yield a negative number, so it doesn't get rejected in the domain of the [imath]\ln[/imath] function, why the final answer just give the answer like this [imath]\cos^{-1}(z) = -i\ln(z+\sqrt{z^2-1})[/imath]???

You said that cos^-1(z) won't yield a negative number. Of course it can! What you meant to say was that the argument of ln (that is what you are taking the ln of) can't be negative. As Dr Peterson pointed out, even that you should not be concerned about that for the most part.