Sign up now
to enroll in courses, follow best educators, interact with the community and track your progress.
Download
Modulus and Argument of a Complex Number
5,349 plays

More
This lesson explains the concept of Modulus and argument of a complex number for JEE Main and Advanced

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE | 6 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
Unacademy user
AA
Hello Sir ! , I have enrolled on all of your courses for Prelims 2018, they are very well made and explained. Sir I wanted to know whether all the courses - Geography, Economics and Polity would be completed by 15th May 2018, so that we can get sufficient time for revision ? Thank You .
Chetan Gaurav
2 years ago
Hi Ayush , Thank you so much enrolment :) Yes even before that it will be completed:) In Indian Economy- Major major topics are already finished... With that much knowledge you will be able to solve the questions but one portion which is still left is budget, economic survey & economy related schemes by the GOI...For that I am planning and I have to check the same with the unacademy team.If something positive comes out than will go for that also definitely. In Polity - I have covered 90% course... and rest will be covered in upcoming or maximum by 20 April ( Including some current affair topics related to polity also as upsc will link the basics with current issues ..Will not go Much in detail but will study that from examination point of view) Geography- Will be covered by the end of this month hopefully if everything works out as planned :)
AA
Ayush Agarwal
2 years ago
Great ! Thank You so much Sir, Your way of making notes and mentioning relevant exam related information have saved so much of my time , I would definitely recommend your course to other UPSC Aspirant. :)
sir ji aapki slide mai 2nd property mei ek mistake hai...vo |z1 - z2| vali mai.. bass baaki tohh best lecture tha.. loved it
Vineet Loomba
7 months ago
Kya mistake h dost ?
Pratham M
7 months ago
sir ji |z1 - z2| ≥ | |z1| - |z2| | sirf whole Mai modules Nahi laga hua tha
sir i meed 1 st year lessons also for preparation of emcet and mains
Vineet Loomba
7 months ago
All avlbl on my profile
Sir I have a doubt on mathematical reasoning iit mains 2018
Vineet Loomba
10 months ago
Wht doubt ?? ..see the course on Mathematical reasoning
Avinash Kumar
10 months ago
Just a question of 2018 mains i am asking otherwise i know the chapter
sir please , i need course on probability . jaldi bana dijiye sir.
Vineet Loomba
a year ago
it will take time as this is covered in last.
Rangnath Gote
a year ago
ok
Thank you sir for this excellent course, I feared complex before this course but you made it look easy. Thank you Sir, keep adding more such useful courses. Just one request -- add Permutations and combinations.
Vineet Loomba
2 years ago
most welcome ...will add soon
  1. IIT-JEE CRASH COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Search vineet loomba unacademy" on GOOGLE IIT-JEE MATHS REVISION COURSE PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENtoR


  2. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010. Doubts/Feedback/Ideas in Comment Section. Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates. Share among your peers as SHARING is CARING !!


  3. COMPLEx NUMBERS FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) #MODULUS AND ARGUMENT IIT-IEE MATHS REVISION COURSE PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR


  4. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE MoDULUS OF A COMPLEX NUMBER The modulus, which can be interchangeably represented by orr, is the distance of the pointz from the orign, sothat its numerical value isgiven by _r_Vr'T MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  5. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE Example: Find the complex numbers zwhich simultaneously satisfy the equations Solution: z-125 z-4 z-8i3 z-8 x-4+iy Here z-125 x=012-sil-3 -25y +136-0 9(364 y*)-2536 + (y-8 y = 17, 8 Hence the required numbers are z = 6 + 1 71 6 + 8 i. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  6. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE PRINCIPAL ARGUMENT OF A CoMPLEX NuMBER 2 z lies in the first quadrant z lies in the second quadrant MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  7. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE PRINCIPAL ARGUMENT OF A CoMPLEX NuMBER z lies in the third quadrant 2 lies in the fourth quadrant 2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  8. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE Method of finding the principal value of the argument ofa complex number z = x + y. Step 1 : Find tan and this gives the value of in the first quadrant. StepII: Find the quadrant in which z lies, with the help ofsign of x and y coordinates. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  9. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE b> 0, then + b > 0, then (i) If z lies in the I quadrant principal argument-6- and general argument-2 k If z lies in the II quadrant principal argument-6-7- and general argument-2 If z lies in the III quadrant principal argument-0- + and general argument-2 k + ( - ) If z lies in the IV quadrant principal argument-0- and general argument-2 k- i.e. a >0 and (ii) i.e. a < 0 and + ( - ) (iii) i.e. a < 0 and b< 0, then (iv) i.e. a> 0 and b < 0, then MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  10. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE POLAR FORM z r (cos + i sin ) EULeR FORM z=re =cos + i sin MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  11. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE PROPERTIES arg (zi . Z2 3 z") = arg (zi) + arg (z,) + + arg (z.) 2.121-211 arg()=arg (zi )-arg (:) . 4 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  12. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE Example: If z-2il 2, then find the greatest and least of z Solution: we have lz _ 2 + il -12-11-11 -,51 Hence greatest value of lis 5 + 2 and least value of lis 5-2. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)