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Finding all values in C of arcsech(i) watch

1. Hello, I was wondering if anyone could check my solution for this problem, The solution follows as:

First note that . So , this means that we have two cases due to the so taking the positive case first we have that , now recall that in this case which implies that and since there is no real component of and the imaginary part is greater than 0, so we must have that where .
My main concern is the argument i calculated i'm unsure if its correct , i assume if my case is correct for the positive then it will be the same method for the negative. Any help would be highly appreciated.
2. Pricipal argument:
Yes, your principal argument of is fine as is purely imaginary and is of form ai, where .

As a check, try , where is the answer in your first post.

Though for part of your last equation, instead of , you need .

Method for negative second case:
Yes, similar method for the second negative case.

For the real case, as is an even function, the positive case is usually used of so to return positive values (depending on context of course).
3. (Original post by simon0)
Pricipal argument:
Yes, your principal argument of is fine as is purely imaginary and is of form ai, where .

As a check, try , where is the answer in your first post.

Though for part of your last equation, instead of , you need .

Method for negative second case:
Yes, similar method for the second negative case.

For the real case, as is an even function, the positive case is usually used of so to return positive values (depending on context of course).
Hi thanks for your response I understand most of what you are saying I just don't see how there can be a real case
4. (Original post by simon0)
Pricipal argument:
Yes, your principal argument of is fine as is purely imaginary and is of form ai, where .

As a check, try , where is the answer in your first post.

Though for part of your last equation, instead of , you need .

Method for negative second case:
Yes, similar method for the second negative case.

For the real case, as is an even function, the positive case is usually used of so to return positive values (depending on context of course).
Also what would the Argand diagram look like for the positive case ?
5. (Original post by Scary)
Hi thanks for your response I understand most of what you are saying I just don't see how there can be a real case
I meant of where .
6. (Original post by Scary)
Also what would the Argand diagram look like for the positive case ?
Your solutions for the positive case are of form:

where .

This is just a complex value (of form a + bi) and using the pricipal argument, can you now see how we can create an argand diagram?
7. (Original post by simon0)
Your solutions for the positive case are of form:

where .

This is just a complex value (of form a + bi) and using the pricipal argument, can you now see how we can create an argand diagram?
Yeah the a value is constant so it will just be a set of solutions on that vertical line repeating up and down the imaginary axis
8. (Original post by Scary)
Yeah the a value is constant so it will just be a set of solutions on that vertical line repeating up and down the imaginary axis
Yes and of course the set of solutions for the negative case.
9. (Original post by simon0)
Yes and of course the set of solutions for the negative case.
Brilliant, thank you for your help, i was just unsure about the diagrams, also a quick question if we had and wanted to find all of it's values in when we plot an argand diagram for these values just all lie on the negative real axis? using the fact that using this i calculated .
10. (Original post by Scary)
Brilliant, thank you for your help, i was just unsure about the diagrams, also a quick question if we had and wanted to find all of it's values in when we plot an argand diagram for these values just all lie on the negative real axis? using the fact that using this i calculated .
Looks good to me.

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