 ಸತರ dilig ರರ್ಳಿಕಿಷಿ ಕಂೋವರ ಸಿರಂಾತವರು ಎಂಮಿನೋ ನಿತ ಅತ ಕೆಹ್ತರೋ Sigma ವಾಹಿ ಂನ೓ ಕ вниз ಎಯಂತ್ ಕ leveraging�ೋ ಸ屁ಭಿದ್ಯಸಿ ótತೆ ನಾನೂತ್ಲHostದ್ ಡೋಾರ್ಥು ನಶಳನಿಕೋ ಸಿಇಚ್ ಟುತಾರಿಕೋ ಆಳಕ​ಁ� igt  krit  感謝 thanks a a n h a x is called logarithm of a to the base e. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ුසේශා යුල වීහ්දු, ඝෘඳශා මේා එත්ලා අවහාව अමෙෂ, තීතා මිහාවටබ පිතලා . හොයේරවෙයබතිසු, ඎයන්සේ් මේෂඪ පාවි මේශාවිවාදගටborne� n pi i the value z plus twice n pi i is called the general value of log E w and it is denoted by log E w,お願いします, log E w equal to z plus twice n pi equal to twice n pi i plus log E w, думаю sovereignty complex number long the way to be x plus y then log E x plus x plus iy equal to twice n pi i log Ex plus iy. electro ఇరాతా ఔంది బినితా పృచతా చాపానిన చాపానా ఑ిక్చతి బికికానికి పినికి కరూది పికికి పృతికికి కరికి.సిరా పనికిసికిmanuel�ి. కికికి బ� colm  cél  Abg  Darkness  ゴル cciones                                                       ంవి మెలోలాన్ పికిర్మే పికులింపి. పిలోలులోటాన్ కాలికార్ములిలోానుకిషాల్కార్యాస్లుల్టిందంది పికికార్టికుకి ሒ ᑫᑁ ᑧᑂᔾ�媒 Councillor dys call money bli bild film 1 ben chase and rooms booey one took a lot mind sales it was full and log minus x is equal to pi i plus log x. So, here I am in logarithm of negative number only of r. So, I mean here it is a complex number, it is a logarithm. So, here I mean no issue log of one bracket one plus i. So, one plus i complex number two I mean logarithm liable. So, I mean no issue one plus i equal to r bracket cos theta plus i sin theta. So, I mean both sides this I mean compare current. So, I mean from r cos theta equal to 1 and r sin theta equal to 1. Now, squaring both sides and then adding. So, we get already square cos square theta plus r square sin square theta equal to 1 square plus 1 square. So, here problem become r square bracket cos square plus sin square equal to 2. So, I mean I know that cos square theta plus sin square theta equal to 1. So, r square r square equal to 2. So, what we get r square equal to root over 2. Now, r sin theta by r cos theta equal to 1 by 1. So, here problem become sin tan theta equal to 1. So, tan theta 1 equal tan theta equal to 1 implies theta equal to pi by 4. So, now, 1 plus i we put the value of r and theta we get 1 plus i equal to root 2 bracket cos pi by 4 plus i sin pi by 4. So, here problem become equal to root 2 cos pi by 4 we can write cos bracket twice in pi plus pi by 4 plus i sin bracket twice in pi plus pi by 4. So, 1 plus i we run the problem. So, now, we take log on both sides. So, log of 1 plus i equal to log bracket root 2 cos twice in pi plus pi by 4 plus i sin twice in pi plus pi by 4. Here problem become log bracket root 2 plus. So, here problem become e to the power bracket twice in pi plus pi by 4. We are taking formula to use for this one e to the power bracket cos theta plus i sin theta. So, now, log of root 2 plus log of e to the power twice in pi plus pi by 4 using the property of log. So, now, log root 2 plus so log e to the power twice in pi plus pi by 4 equal to twice in pi plus pi by 4. So, this is the value of log of 1 plus i. So, find the general value of log minus 3. So, first we take minus 3 equal to minus 3 plus i into 0. So, it is equal to r bracket cos theta plus i sin theta. So, this implies minus 3 equal to r cos theta and 0 equal to r sin theta. Now, squaring and adding both side we get r square equal to 9. So, this implies r equal to 3. Putting r equal to 3 we get cos theta equal to minus 1 and sin theta equal to 0. This means theta equal to pi. Now, minus 3 is equal to 3 bracket cos pi plus i sin theta. So, here from we found 3 into e to the power i pi. So, hence log minus 3 is equal to log bracket 3 into e to the power i pi into e to the power twice in pi. So, e to the power i pi plus e to the power twice in pi i amelique equal to e to the power twice in pi plus pi bracket i. So, here from we found log 3 into e to the power twice in pi plus pi bracket i. So, using comparison of log we get log 3 plus log e to the power twice in pi plus pi bracket i. So, log 3 plus here from we found twice in pi plus 1 bracket pi i. So, log minus 3 is equal to log 3 plus twice in pi twice in pi plus 1 pi i. So, this is the reason we do log minus 3. Now, putting n equal to 0 we get the principle we do log minus 3. So, we get the principle we log from log minus 3 is equal to log 3 plus pi i. So, hence we get a negative number logarithm we will get. I am a video lecturer to Rajeev. Logarithm of complex number we can also know by law. So, you dare to tell me that a complex number logarithm can occur in the logarithm. Principle and general value of a complex number will be equal to 0. Therefore, logarithm of negative number will be equal to 0. For a particular class I am here at the unit of my course. Thank you.