this is what the ideal content/text in a valentine card should be
If someone can verify this, you will be my valentine.
i = imaginary number i
^ or exp = exponent
pi = pi
arg = argument
find a value for [e/2(-1-iroot3)]exp 3(pi)i
[my soln]
[e/2(-1-iroot3)]exp 3(pi)i
= e^(3pi)ilog[e/2(-1-iroot3)]
letting z = [e/2(-1-iroot3)]
log(z) = ln|z| + iarg(z)
from [e/2(-1-iroot3)], |z|= e, arg(z) = 4pi/3
therefore,
log(z) = ln(e) + i (4pi/3) = 1 + i (4pi/3)
hence, (3pi)ilog(z) = (3pi)i -4(pi)^2
THEREFORE a solution to
[e/2(-1-iroot3)]exp (3pi)i = e^(3pi)i -4(pi)^2
----
When I came out of the quiz for complex variables, I was kicking myself in the ass because I thought ln(e) was equal to zero, but on the quiz I wrote 1. After verification at home that ln(e) was in fact equal to ONE! then I might have a chance at getting the question right! ...That is, if the above was correct.
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I am currently "writing" my PBL mini review. I have nothing written down... I'm so not sleeping tonight.
i = imaginary number i
^ or exp = exponent
pi = pi
arg = argument
find a value for [e/2(-1-iroot3)]exp 3(pi)i
[my soln]
[e/2(-1-iroot3)]exp 3(pi)i
= e^(3pi)ilog[e/2(-1-iroot3)]
letting z = [e/2(-1-iroot3)]
log(z) = ln|z| + iarg(z)
from [e/2(-1-iroot3)], |z|= e, arg(z) = 4pi/3
therefore,
log(z) = ln(e) + i (4pi/3) = 1 + i (4pi/3)
hence, (3pi)ilog(z) = (3pi)i -4(pi)^2
THEREFORE a solution to
[e/2(-1-iroot3)]exp (3pi)i = e^(3pi)i -4(pi)^2
----
When I came out of the quiz for complex variables, I was kicking myself in the ass because I thought ln(e) was equal to zero, but on the quiz I wrote 1. After verification at home that ln(e) was in fact equal to ONE! then I might have a chance at getting the question right! ...That is, if the above was correct.
------
I am currently "writing" my PBL mini review. I have nothing written down... I'm so not sleeping tonight.