What is the complex logarithm of i in rectangular form?

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Homework Help Overview

The discussion revolves around finding the complex logarithm of the imaginary unit \(i\) expressed in rectangular form \(z = x + iy\). Participants explore the properties of complex logarithms, particularly focusing on the multi-valued nature of the logarithm and the use of Euler's formula.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using Euler's relation to express \(i\) and question the validity of their approaches. Some suggest expressing \(i\) in exponential form and identifying its magnitude and argument. Others highlight the need to consider both the principal and general values of the logarithm.

Discussion Status

The discussion is active with various interpretations being explored. Some participants provide guidance on the correct approach to expressing the logarithm, while others emphasize the importance of recognizing the multi-valued nature of the logarithm. There is no explicit consensus on a single method, but productive direction is evident.

Contextual Notes

Participants note that the use of \( \ln \) typically refers to the principal value of the logarithm, and there is an ongoing discussion about the implications of choosing different branches for the argument of the logarithm.

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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 e^{i\pi/2}=i. What is the log of i then?
Note: Can you confirm that e^{i\pi/2}=i and not e^{i\pi/2}=-i?
 
Millennial said:
One thing to note is that e^{i\pi/2}=i. What is the log of i then?
More generally, i= e^{i(\pi/2+ 2k\pi)} for any integer k (thus giving what Curious3141 called the "general form").

Note: Can you confirm that e^{i\pi/2}=i and not e^{i\pi/2}=-i?
 
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.
<br /> i = e^{i \frac{\pi}{2}}<br />
 
HallsofIvy said:
More generally, i= e^{i(\pi/2+ 2k\pi)} 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 \log and not ln except when you're referring to the natural log of a real number:

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

and where \theta 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 2n\pi i part is to account for the multivaluedness of log but if you just want the principal value, set n=0.
 

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