IIT JEE , AIEEE Forum - Mathematics , Trignometry

madhuvanreddy

logi=

Aatish

we can write

i = cos (pi/2) + i sin ( pi/2 ) = e^i(pi/2)

therefore

log i = log_e ( e^i(pi/2) )

= i (pi/2)

cheeeeerrsssssssss!!!!!!!!

Aatish

any other approach???

iitkgp_bipin

Good approach. Answer is i.(pi/2)

By the way remember this general formula :

ln(a + ib) = (1/2)ln(a² + b²) + i.tan^-1(b/a)

hsbhatt

Adding to what Bipinji is saying

ln(a+ib) = 2n

i + (1/2)ln(a² + b²) + i.tan^-1(b/a) is the more general expression.

It may puzzle you that if we put b = 0, i.e. for a real number, you get infinite values for its logarithm of a real number and that too in complex numbers. The explanation is quite simple.

Remember the polynomial expansion or more appropriately, the Taylor expansion of ln(1+x) = x - x²/2 + x³/3 -.....

That's a polynomial with infinte degree. So infinite solutions in complex numbers.

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