Solving for ln(-e), (-1)^i, and (2i)^1+i in Rectangular Form

[ilikephysics]

Find the value of ln(-e), (-1)^i, and (2i)^1+i in rectangular x+iy form, using the principal value of the logarithm.

This is what I'm doing for each one:

ln(-e)= e^ln(-e)

(-1)^i= e^(i)(ln(-1))

(2i)^1+i= e^(1+i)ln(2i)

If this is right, where do I go from here?

 

[HallsofIvy]

Well, each statement is TRUE but do they help at all?

One reason the problem specifically says "in rectangular form" is that you will want to convert to "polar form" to do the calculations- then convert back.

If you write a complex number z= x+iy in "polar form" as
r e^iθ, then ln(z)= ln(r)+ iθ Notice that in the first for, r e^iθ, adding or subracting π to θ will not change z but will change ln(z). The point of "principal value of the logarithm" is to specify θ: it must be between 0 and 2π.
In particular, -e= e e^i(π). That is, r= e and θ= iπ. ln(-e)= ln(e)+iπ= 1+ iπ

For (-1)ⁱ, use the fact that (e^iθ)ⁱ= e^(i*i)θ= e^-θ. -1= e^iπ so -1ⁱ= e^-π.

 

[M98Ranger]

To solve for ln(-e), we can use the properties of logarithms to rewrite it as ln(e^-1) and then use the fact that ln(e) = 1 to get ln(-e) = -1. This means that the rectangular form of ln(-e) is -1 + 0i.

For (-1)^i, we can use the fact that i is equal to ln(-1)/ln(e) and rewrite it as e^(ln(-1)/ln(e)). Using the properties of logarithms again, we can rewrite this as e^(ln(-1))^(1/ln(e)). Since ln(-1) is equal to i*pi, we have (-1)^i = e^(i*pi)^(1/ln(e)). We can then use Euler's formula, e^(ix) = cos(x) + i*sin(x), to get (-1)^i = cos(pi/ln(e)) + i*sin(pi/ln(e)). Therefore, the rectangular form of (-1)^i is cos(pi/ln(e)) + i*sin(pi/ln(e)).

Finally, for (2i)^1+i, we can use the properties of logarithms once again to rewrite it as (2i)^ln(2i)^(1/ln(e)). Using Euler's formula, we can rewrite this as (2i)^(i*ln(2))^(1/ln(e)). Simplifying, we get (2i)^i^(ln(2)/ln(e)). Again, using Euler's formula, we can rewrite this as (cos(ln(2)) + i*sin(ln(2)))^(ln(2)/ln(e)). Finally, using De Moivre's theorem, we can simplify this to (cos(ln(2)*ln(2)/ln(e)) + i*sin(ln(2)*ln(2)/ln(e))). Therefore, the rectangular form of (2i)^1+i is cos(ln(2)*ln(2)/ln(e)) + i*sin(ln(2)*ln(2)/ln(e)).