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Ln(i) in rectangular form

  • Thread starter leonmate
  • Start date
  • #1
84
1
Express in the following form z = x + iy:

q. ln(i)

This question came up in an exam i had today.. dont think i answered it correctly though!

i went for the euler relation:

e^i*θ = cos(θ) + i*sin(θ)

next i set θ = 1 and found the natural log of the equation twice

ln(i) = ln(ln(cos(1) + i*sin(1)))

My calculator wasn't good enough to work this out without returning a math error, so i left it in that form (except i think i calculated the trig functions)

Can anyone shine some light on this problem?
 

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
86
Express in the following form z = x + iy:

q. ln(i)

This question came up in an exam i had today.. dont think i answered it correctly though!

i went for the euler relation:

e^i*θ = cos(θ) + i*sin(θ)

next i set θ = 1 and found the natural log of the equation twice

ln(i) = ln(ln(cos(1) + i*sin(1)))

My calculator wasn't good enough to work this out without returning a math error, so i left it in that form (except i think i calculated the trig functions)

Can anyone shine some light on this problem?
You're using the wrong approach.

First, recognise that the complex logarithm is a multi-valued function. They might be asking you to find the principal value, or the general value. For completeness, you should state both.

Start by expressing "i" itself in exponential form. What are the magnitude and argument of i?
 
  • #3
296
0
One thing to note is that [itex]e^{i\pi/2}=i[/itex]. What is the log of i then?
Note: Can you confirm that [itex]e^{i\pi/2}=i[/itex] and not [itex]e^{i\pi/2}=-i[/itex]?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,770
911
One thing to note is that [itex]e^{i\pi/2}=i[/itex]. What is the log of i then?
More generally, [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for any integer k (thus giving what Curious3141 called the "general form").

Note: Can you confirm that [itex]e^{i\pi/2}=i[/itex] and not [itex]e^{i\pi/2}=-i[/itex]?
 
  • #5
Bacle2
Science Advisor
1,089
10
Actually, a helpful way of looking at logz is that it gives you the coordinates

of the point z --radius and argument . But, since, as curious said, logz is multivalued,

you need to select a branch in which to define the argument. If you choose, e.g.,

(-Pi,Pi) , your argument is the angle (starting at Pi) between the positive x-axis and

z=i.
 
  • #6
2,967
5
When ln (or Log) is used, instead of log, it is understood that the principal value of the logarithm is given.
[tex]
i = e^{i \frac{\pi}{2}}
[/tex]
 
  • #7
296
0
More generally, [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for any integer k (thus giving what Curious3141 called the "general form").
My statement still holds. In fact, I wanted the OP to understand that on his own.
 
  • #8
1,796
53
Express in the following form z = x + iy:

q. ln(i)

This question came up in an exam i had today.. dont think i answered it correctly though!

i went for the euler relation:

e^i*θ = cos(θ) + i*sin(θ)

next i set θ = 1 and found the natural log of the equation twice

ln(i) = ln(ln(cos(1) + i*sin(1)))

My calculator wasn't good enough to work this out without returning a math error, so i left it in that form (except i think i calculated the trig functions)

Can anyone shine some light on this problem?
That is way wrong. Just use the formula for the complex logarithm and use [itex]\log[/itex] and not ln except when you're referring to the natural log of a real number:

[tex]\log(z)=\ln|z]+i(\theta+2n\pi)[/tex]

and where [itex]\theta[/itex] is the principal argument of z. Now, since i on the Argand diagram is located at the point (0,1), that means it's principal argument is pi/2 so there you have it and that [itex]2n\pi i[/itex] part is to account for the multivaluedness of log but if you just want the principal value, set n=0.
 

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