## anonymous 4 years ago Evaluate Log e (2+J5) giving the answer in the form a+jb .

1. dumbcow

1.68 + j1.19

2. anonymous

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

3. dumbcow

yeah thats what i assumed

4. anonymous

so how did you get the answer..

5. dumbcow

haha calculator :)

6. anonymous

ln2+...

7. anonymous

how did you use calculater..

8. anonymous

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

9. dumbcow

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

10. dumbcow

yes i am aware of that

11. anonymous

but it is not i, it is j

12. anonymous

never mind then..

13. dumbcow

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

14. anonymous

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

15. anonymous

i assume this is a complex number

16. anonymous

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

17. anonymous

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

18. anonymous

I like this one

19. anonymous

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

20. anonymous

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

21. anonymous

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