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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.
Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Upasana!!!
Regards,
Priyansh Bajaj
Askiitians Experts
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