Question

find the value of logi ? explain it​

Answer 1

Answer:

Let a+ib=sin(logi

i

)=sin(ilogi)

=sin(i(log∣i∣+iamp(i)))…(i=∣i∣amp(i))

=sin(i(0+

2

iπ

))

=sin(0+

2

i

2

π

)

=sin(0−

2

π

)

a+ib=−1

⇒a=−1,b=0

∴sin(logi

i

)=−1+0=−1

Now cos(logi

i

)=

1−sin

2

(logi

i

)

=

1−1

=0