11if i had the authority i would gibe u RED for asking pink[9]
wat is value of sin(log ii)
iiii this power upto infinity = a + ib find them ....find b/a and a2 +b2
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24 Answers
1dude i have a doubt..........
isnt r=√a2+b2????
so wont a2+b2=r2
=e-b∩ ???????
1ohhhhhhhhhhhhhhhhhhhhhhhhhhhh dint see that.......im extremely sorry hehe
11Let Z = a + ib = rei∂
Then, arg(Z) = tan-1(b/a) = ∂
and |Z| = √(a2+b2) = r
...thats all !!
1dudr rk.........i dint understand ur last and second last steps......plzz explain it naa..............i think its conceptual..... i think im lackin sum concept...
11Q2.
i i i i...∞ = a + ib
(eiπ/2)(a + ib) = a + ib
e(-bπ/2 + i(aπ/2)) = a + ib
e(-bπ/2).e(i(aπ/2)) = a + ib
arg(a + ib) = tan-1(b/a) = aπ/2
So, b/a = tan(aπ/2)
|a + ib| = √(a2 + b2) = e(-bπ/2)
So, a2 + b2 = e(-bπ)
1wel i thought u wer typin d solution so i dint reply for one........please giv d solution naa.....
1iii= a +ib
iii lni = ln(a+ib)
=> iii ln(eipi/2) = ln(a+ib)
=>iii.ipi/2 = ln(a+ib)
next wat????????
1ii = x +iy
i lni = ln(x+iy)
=> i ln(eipi/2) = ln(x+iy)
=>i2pi/2 = ln(x+iy)
=> -pi/2 =ln(x+iy)
=> x+iy = e-pi/2
so, x=e-pi/2 and y=0
so, ii = e-pi/2
log ii = -1.
1yup it is -1.............
eiθ=cosθ+isinθ
eiπ/2=i
iπ/2=logi
-Ï€/2=logii
so sinlogii=sin(-Ï€/2)=-1..