Go Back > Great Internet Mersenne Prime Search > Math

Thread Tools
Old 2004-11-05, 03:10   #1
dave_0273's Avatar
Oct 2003
Australia, Brisbane

2×5×47 Posts
Default Complex number problem

I am a maths tutor and there is one question on one of my students assignments that I just can't get the answer for.

I have to express ln(2*(3)^(1/2) - 2i) in standard form (a + bi).

(it is hard to read, so it is the natural log ( 2*root3 -2i) if that makes it easier to read)

Any help would be appreciated. I have done all of the questions that are similar but I just can't seem to get this one out.

Thanks in advance for you help.
dave_0273 is offline   Reply With Quote
Old 2004-11-05, 03:31   #2
Dec 2003

3 Posts

Set the expression you gave equal to some complex variable, say c+di, then exponentiate both sides (e.g. e^z), and go from there; hope that helps.
Starrise is offline   Reply With Quote
Old 2004-11-05, 04:41   #3
jinydu's Avatar
Dec 2003
Hopefully Near M48

2×3×293 Posts

In general, the natural logarithm of a number gives an infinite number of answers. According to

ln z = ln r + (t + 2npi)*i

where r is the absolute value and t is the argument.

Now, let's look at your expression:

ln(2sqrt(3) - 2i)

r = sqrt [ (2sqrt3)^2 + (-2)^2 ]
r = sqrt [ 12 + 4 ]
r = sqrt [16]
r = 4

t = arctan ( (-2) / (2sqrt(3)) )
t = arctan ( -1/(sqrt(3)) )
t = 5pi/6

If we let z = ln(2sqrt(3) - 2i), then using the formula, we have:

z = (ln 4) + (5pi/6 + 2npi)*i
jinydu is offline   Reply With Quote
Old 2004-11-08, 17:15   #4
Bronze Medalist
mfgoode's Avatar
Jan 2004

80416 Posts
Lightbulb Complex number problem

[Any help would be appreciated. I have done all of the questions that are similar but I just can't seem to get this one out.

Thanks in advance for you help.[/QUOTE]

Dave Im not sure whether you only require the logarithm (Log) in (a+bi) form and that should suffice or you want to evaluate the given log as Jinydu has done. I dont like to approve the use of a formula when its derivation is not given or known esp. in the case of tutors and teaching

I give below your log in a+bi form from 1st principles but can also derive the formula used by Jinydu if required.

To make the derivation more easily readable I have dispensed with the sign of multiplication * unless it is absolutely necessary to avoid ambiguity. Also Instead of writing 'pi' I write 'p'. Thus pqr means pi*q*r

Let the given log (2sq.rt.3-2i) be denoted by x+yi where x=2sq rt.3 and
Y=-2i Here i=sq rt.-1
Let log(xi+yi)=(a+bi)
Therefore x+yi =e^(a+bi)-------- def of logs ---(1)
Now it is well known (after Euler) that for all integral values of n
e^2npi =[ cos 2np +i sin 2np] =1 ----(2)
Now x + yi =e^ (a + bi) from (1)
i.e. x+yi =e^ (a + (bi*e^2npi)i-----from---(2) as e^2npi =1
Hence x +yi = e^ ( a +(b +2np)i
So we see that a +bi be a log of x +yi so also is
(a + bi + 2npi)
i.e. a + (b +2np)i
This is the form of [a + Bi] where B = (b +2np) which is required

Hope this is clear enough. If you require the proof of the formula used by Jinydu It can be given. If you want to try it yourself convert (2) into polar co-ordinates and you will get

Log ( a +bi ) = log sq.rt. (a^2 + b^2) +i (2np + arc tan( b/a))

mfgoode is offline   Reply With Quote

Thread Tools

Similar Threads
Thread Thread Starter Forum Replies Last Post
the unit circle, tangens and the complex plane bhelmes Math 0 2018-03-04 18:50
Basic Number Theory 10: complex numbers and Gaussian integers Nick Number Theory Discussion Group 8 2016-12-07 01:16
Faking factors with Complex Multiplication ATH GMP-ECM 14 2006-12-04 23:48
New Number Pyramid Logic Problem edorajh Puzzles 2 2004-01-14 09:53
A problem of number theory hyh1048576 Puzzles 0 2003-09-28 15:35

All times are UTC. The time now is 10:48.

Thu Dec 2 10:48:20 UTC 2021 up 132 days, 5:17, 0 users, load averages: 1.09, 1.19, 1.24

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.