What is the complex logarithm of i in rectangular form?

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Express in the following form z = x + iy:

q. ln(i)

This question came up in an exam i had today.. don't 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?
 
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leonmate said:
Express in the following form z = x + iy:

q. ln(i)

This question came up in an exam i had today.. don't 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?
 
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]?
 
Millennial said:
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]?
 
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.
 
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]
 
HallsofIvy said:
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.
 
leonmate said:
Express in the following form z = x + iy:

q. ln(i)

This question came up in an exam i had today.. don't 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.