Complex number problem

[dave_0273]

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.

[Starrise]

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.

[jinydu]

In general, the natural logarithm of a number gives an infinite number of answers. According to http://mathworld.wolfram.com/NaturalLogarithm.html

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

[mfgoode]

[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))

Mally