IIT JEE , AIEEE Forum - Mathematics , Algebra

umang

Que. Find the values of :

1. log i

2. iⁱ

Note : i stands for iota

kiran

Hi

Let z be a complex number

log z = log |z| + i arg(z)

= log |z| +i theta.

Here z = 0+ i |z| = 1 and arg(z) = pi /2

Hence log i = log 1 + i*pi/2 = i*pi/2

For i^ i

i^i = e^( i*log i )= e^ (i * log (1+i*(pi/2 + 2pi k) ) = e^ - pi/2

Hope its clear .

Cheers!

vasanth

i = e^i/2---------------------------------euler form

therefore logi = loge^i/2

= i

/2

kiran ur answer is wrong!!!!!!!!!!

------------------------------------------------------------------------------------------------------

i^i =e^ilogi= e^{i i/2}[ log i = i

/2------from previous problem]

= e^-/2

hope u got it

if u found ma answer short and useful

then plzzzzz rate me

cheeeeeers

vasanth

hey

dont mind to ask if u're unclear

always ready 2 explain

cheeeeers

kiran

Hi friends

Actually Vasanth my answer is correct. It was a printing mistake . I had written 1 instead of log 1. So the answer of log i = i*pi/2 .

The method I have given is applicable to all complex numbers. So You can Use it anywhere.

Cheers!

umang

Hey vasanth !

Thanks for the solution !!!!

And kiran , pls explain how you got this formula -

log z = log |z| + i arg(z)

Thank you both !!!!!

miyo

Hey umang,

u know z = lzl*cis

; { cis

= eⁱ ; arg(z) =

} ;

Apply log on both sides.....

vasanth

yeah miyo's rit

z = |z| .eⁱ

log z = log |z| + log eⁱ

just one more step and u get ur answer

cheeeeeeeers

t_c511

hey guys

iⁱ has infinite solutions e^-/2 is one of its solutions
i = cos( 2k

+

/2) + i sin( 2k

+

/2) k

integers

hence for different values of k we will get same value
and hence iⁱ has many answers
this is common mistake made by every1

reply if any doubts

vasanth

alrite just a small modification

i^i = e^{i logi}

=e^{i i(2kpi +pi/2)}

= e^-(2kpi+pi/2)

thanx 4 pointin out ma mistake

i'll giv u a salute

cheeeeeeers(plzzzzzzz end ma rate drought)

nishanth61

coooooooooooool

i'll give u a rate

sanjeev_mishra

let logi=x

e^x_=i

^_e4x_=1=e0

^4x=0

^x=logi=0

i^i_=-1

umang

yes , u all are right !!!!!

Thanks a ton !

vasanth

the answer to the first q can b written as

logi = i(2kpi +pi/2)

-------------------------------------------------------------------------------------------------------------------

hey sanjeev ur steps r understandable

but u c in the penultimate step

u've got

log i =0

which means e^0 =i

which is not possible

so i'd b satisfied if u could clear this doubt o' mine.......

by the way plzzzzzz explain how u got i^i = -1

cheeeeers

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