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

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


1 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
11 years ago

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¯ =

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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All the best Upasana!!!


Priyansh Bajaj

Askiitians Experts

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