For mathlovers please help me In (3+4i)= ?????

In (3+4i)= ???????????????????????????

please helpp meeeeeeeeeee

First, convert (3 + 4i) into polar form.
r = √(3² + 4²) = 5
θ = arctan(4/3) = 0.9273

(3 + 4i) = 5(cosθ + i sinθ) ; where θ =arctan(4/3)
(3 + 4i) = 5e^(iθ)

ln [3 + 4i] = ln [5e^(iθ)]
ln [3 + 4i] = ln 5 + ln [e^(iθ)]
ln [3 + 4i] = ln 5 + iθ
ln [3 + 4i] = ln 5 + i arctan(4/3)
ln [3 + 4i] = ln 5 + 0.9273 i

Did you know that you can google

ln(3 + 4i) and it returns the following from google's calculator?

ln(3 + (4 * i)) = 1.60943791 + 0.927295218 i

i is the square root of -1. It is a number that has no value because there is no number that can be multiplied by itself to equal -1. -1 * -1 = +1....Anyway, in most cases you can't remove the imaginary component from expressions once it appears.

Do you want ln(3+4i) as in natural logarithm? If so
ln(a+bi) = (1/2)ln(a^2+b^2) + i*tan-1(b/a)
So ln(3+4i) = (1/2)ln25 + itan-1(4/3)
= ln5 + i*tan-1(4/3) = 1.6095 + .9273i

ln (3 + 4i)
= 1.6094 ... + 0.927 ... i

very easy:

remember e^lnx=x

so

e^ln(3+4i)=3+4i

i = -1
so 3 + (-4) = -1

ln(5) + i arctan(4/3)

well... thats completely factored out, i dont see exatly what else you need to solve about it