- #1

#### member 508213

If e^(i2(pi))=1 then why can't you take the ln of both sides and have

i2pi=0

?

i2pi=0

?

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- Thread starter member 508213
- Start date

- #1

If e^(i2(pi))=1 then why can't you take the ln of both sides and have

i2pi=0

?

i2pi=0

?

- #2

Homework Helper

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$$e^{2\pi \, i}=e^{0}$$

we cannot conclude in general from

$$\mathrm{f}(x)=\mathrm{f}(y)$$

that

x=y

consider the function f(x)=11

f(67)=f(14.4)

but

67 is not equal to 11

(-3)^4=3^4

but

-3 is not equal to 3

we cannot conclude in general from

$$\mathrm{f}(x)=\mathrm{f}(y)$$

that

x=y

consider the function f(x)=11

f(67)=f(14.4)

but

67 is not equal to 11

(-3)^4=3^4

but

-3 is not equal to 3

Last edited:

- #3

what you say makes sense but I just thought that I would still be able to use the natural log and still keep the statements equivalent, but I do not understand complex numbers very deeply so I will just accept that what I did is not true.$$e^{2\pi \, i}=e^{0\pi \, i}$$

we cannot conclude in general from

$$\mathrm{f}(x)=\mathrm{f}(y)$$

that

x=y

consider the function f(x)=11

f(67)=f(14.4)

but

67 is not equal to 11

(-3)^4=3^4

but

-3 is not equal to 3

- #4

- 346

- 48

If e^(i2(pi))=1 then why can't you take the ln of both sides and have

i2pi=0

?

When you take the ln of both sides, you're trying to invert the exponential function.

The problem is that the complex exponential function is not injective (its not one to one). In order for a function [itex]f\left(x\right)=y[/itex] to have an inverse each x value must correspond to only one y value, and each y value must correspond to only one x value.

The complex exponential is not one-to-one. This is because there are y values that correspond to multiple x values. In your example [itex]y=1[/itex] corresponds to [itex]x=0[/itex] and [itex]x=2\pi i[/itex].

A similar thing happens with the function [itex]y=x^2 [/itex]. Again here each nonzero y value corresponds to two x values and a true inverse does not exist. You can take the square root to find the magnitude of x. But then you have to decide if you want the positive or negative root.

You can run into troubles if you take the the wrong root.

For instance consider the equation [itex]x^2 =4 [/itex]. One solution to this equation is [itex]x =-2 [/itex]. This means that I can write the equation as [itex]\left(-2\right)^2 =4 [/itex]. I could then take the square root of both sides giving me [itex]-2 =2 [/itex]. Obviously this is wrong. The problem is that I was very sloppy in how took the square root of (inverted) the function [itex]x^2[/itex]. Its the same problem you run into when taking the ln of a complex exponential.

- #5

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