# Find all the values of a complex?Watch

Announcements
Thread starter 7 months ago
#1
I'm struggling to find all the values of this number shown below. The question just states find all the values of this:

Attachment 730648

Thank you for any help

The number is:

ln[16e^(-2*pi*i/17)]
Last edited by CRD_98; 7 months ago
0
quote
7 months ago
#2
0
quote
Thread starter 7 months ago
#3
(Original post by Charliewiz)
Thank you, I've re-loaded it and typed it out just in case it doesn't work again
0
quote
7 months ago
#4
(Original post by CRD_98)
Thank you, I've re-loaded it and typed it out just in case it doesn't work again
There's one "obvious" answer.

Bu then you need to think about the different possible arguments the number you are taking the logarithm of
0
quote
Thread starter 7 months ago
#5
(Original post by RichE)
There's one "obvious" answer.

Bu then you need to think about the different possible arguments the number you are taking the logarithm of

Could this be done using ln(z) = ln|z| + iArg(z) ?
Last edited by CRD_98; 7 months ago
0
quote
7 months ago
#6
(Original post by CRD_98)
Could this be done using ln(z) = ln(z) + iArg(z) ?
Yes, though you mean

ln(z) = ln|z| + iArg(z)
0
quote
Thread starter 7 months ago
#7
(Original post by RichE)
Yes, though you mean

ln(z) = ln|z| + iArg(z)
yes thank you, I tried to correct that as soon as a I posted it. Am I right in thinking since z = re^(i.theta) => z = 16 and theta = -2*pi/17

=> ln[16e^(-2i.pi/17)} = ln|16| + i(-2.pi/17 + 2.pi.k) , where kEZ
0
quote
7 months ago
#8
(Original post by CRD_98)
yes thank you, I tried to correct that as soon as a I posted it. Am I right in thinking since z = re^(i.theta) => z = 16 and theta = -2*pi/17

=> ln[16e^(-2i.pi/17)} = ln|16| + i(-2.pi/17 + 2.pi.k) , where kEZ
Yep.

No need for |16|. Just 16 will do.

And could write ln16 = 4ln2 if you wanted.
1
quote
Thread starter 7 months ago
#9
(Original post by RichE)
Yep.

No need for |16|. Just 16 will do.

And could write ln16 = 4ln2 if you wanted.
That’s great, thank your for all your help
Posted on the TSR App. Download from Apple or Google Play
0
quote
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Sheffield Hallam University
Wed, 17 Oct '18
• Staffordshire University
Nursing and Midwifery Undergraduate
Wed, 17 Oct '18
• Teesside University
Wed, 17 Oct '18

### Poll

Join the discussion

#### Who is most responsible for your success at university

Mostly me (52)
98.11%
Mostly my university including my lecturers/tutors (1)
1.89%