Question

Evaluate Log e (2+J5) giving the answer in the form a+jb .

Answer 1

1.68 + j1.19

Answer 2

you looking for this?\[ \ln(2+J5)\]

Answer 3

yeah thats what i assumed

Answer 4

so how did you get the answer..

Answer 5

haha calculator :)

Answer 6

ln2+...

Answer 7

how did you use calculater..

Answer 8

you know that right..
\[\ln(a+b)\neq lna+lnb\]

Answer 9

well you would have to have a graphing calculator (TI 83 or better)
ln(2+5i) , there is symbol for an imaginary number

Answer 10

yes i am aware of that

Answer 11

but it is not i, it is j

Answer 12

never mind then..

Answer 13

same thing, different notation for a complex number which has real part and imaginary part

Answer 14

I've never seen a+jb
http://en.wikipedia.org/wiki/Complex_number

Answer 15

i assume this is a complex number

Answer 16

usually written as
\[\ln(a+bi)\] or more usually
\[\log(a+bi)\]

Answer 17

find it via
\[\log(a+bi)=\ln(|a+bi|)+i\theta\] where
\[\tan(\theta)=\frac{b}{a}\] and
\[|a+bi|=\sqrt{a^2+b^2}\]

Answer 18

I like this one

Answer 19

can we say that then..\[|a+bi|=\sqrt{x^2+b^2}=\sqrt{29}\]
\[\tan^{-1}( 5/2)=68=\theta\]
\[\log(2+5i)=\ln|2+5i|+i \theta\]
=\[\ln \sqrt{29}+i68=1.68+68i\]

Answer 20

actually thisone is better\[\log(a+bi)=\log(|a+bi|)+i\theta\quad or\quad \ln(a+bi)=\ln(|a+bi|)+i\theta\]

Answer 21

68/180=R/pi
\[68 \quad degree \approx \frac25\pi\]
\[\log_{e} (2+5i)=\ln(2+5i)=1.68+\frac25\pi+2k \pi i\]