ei∏ + 1=0;let z=log2 (1+i); then (z+z¯) + i(z+z‾)=?(a) (ln4 + ∏)/ln4(b) (∏ - ln4)/ln/2(c) (ln4 - ∏)/ln4(d) (∏ + ln4)/ln2

Dear Upasana,

Solution:- The general approach for solving such problems is-

1+i = √2 (cos45 + i sin45) = √2 e^(i∏/4)

z = log₂(1+i) = log₂(√2 e^(i∏/4)) = log₂(√2) + log₂(e^(i∏/4))
z = 1/2 + (i∏/4)log₂(e) = 1/2 + i[∏/(4ln2)]

and, z¯ = 1/2 - i[∏/(4ln2)]

z + z¯ = 1 

You must have copied the question incorrectly. From z = log₂(1+i), you can get z = 1/2 + i[∏/(4ln2)] as shown above. Then, you can proceed further. If you face any problems, check the question and post the correct question. 

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Priyansh Bajaj

Askiitians Experts