Super video. The only one I could find that explained how this hangs together rather than just how to do questions. Thank you for providing such an accessible and reliable resource
Great video 👍🏾👍🏾 just wanted to ask if you have arg(w+z) and arg (w) and you had to draw the length z on an Argand diagram could you write arg(w+z-w) according to that formula and get arg(z) therefore making z the half line 👍🏾👍🏾 thank you
I'm not sure how to visualise it geometrically, but the algebra just works. When finding the arg of a complex number one would always do - tan[x] = complex number's opp. (im) length / adj. (ℝe) length. Since we know the complex number itself is (x-3)+i(y+2) and that its arg is -3π/4, the tan equation's angle[x] and im & ℝe lengths can easily be deduced. So you can do a 9:12
inverse tan of O/A, so tan^-1 of (y+2)/(x-3) = -3pi/4, this means that tan(-3pi/4) = (y+2)/(x-3) this defo doesnt help considering you asked a year ago and youve done your exams and youre already 4 months into uni unless you took a gap year so idk
how come u didn't use the distance formula to find the cartesian equation? cuz you're not equating it to another complex no? cuz I did the cartesian equation like in your perpendicular bisector vid and got a different answer
Try looking back at this www.examsolutions.net/tutorials/modulus-argument-complex-number/?board=Edexcel&module=fp1&topic=1745 I hope it helps answer your question
Omg my teacher used your video to give us notes. This is exactly what I have in my exercise book
Super video. The only one I could find that explained how this hangs together rather than just how to do questions. Thank you for providing such an accessible and reliable resource
Thank you, appreciate the your comment.
Incredibly helpful, solid work.
thank you!
This was so much easier to follow than my lecturer. Thank you sir
Exam solutions gives me hope
thanks a million
i had a doubt, but you cleared it in this video. I can't thank you enough.
Pro trick : watch series on Flixzone. Been using them for watching lots of of movies during the lockdown.
@Kameron Nelson Definitely, been using flixzone for since december myself :D
Subscribed, explained much better than my teacher
Thx sir and can you explain demorgan's law also
Great video 👍🏾👍🏾 just wanted to ask if you have arg(w+z) and arg (w) and you had to draw the length z on an Argand diagram could you write arg(w+z-w) according to that formula and get arg(z) therefore making z the half line 👍🏾👍🏾 thank you
Thank you sir!11
God Bless You
ur AMAZING
At 9:25 Im confused on how the tan of the obtuse angle equals opp over adjacent
I'm not sure how to visualise it geometrically, but the algebra just works.
When finding the arg of a complex number one would always do - tan[x] = complex number's opp. (im) length / adj. (ℝe) length.
Since we know the complex number itself is (x-3)+i(y+2) and that its arg is -3π/4, the tan equation's angle[x] and im & ℝe lengths can easily be deduced. So you can do a 9:12
@@ASLUHLUHCE Arctan (y/x) = Arg (z) and if we take tan of both sides, (y/x) = tan9Arg (z)
thanks a lot!!!!
mimisallehh You're welcome
I love you
This is wonderful thank you :) I was just wondering why x < 3 and not x
The start does not contain (3,-2)
If it does, (z-z1)= (0,0), and arg(z-z1) would be undefined.
Ang Jun Liang Ahh, thank you :)
At 9:13 how d'you get to using tan?
inverse tan of O/A, so tan^-1 of (y+2)/(x-3) = -3pi/4, this means that tan(-3pi/4) = (y+2)/(x-3) this defo doesnt help considering you asked a year ago and youve done your exams and youre already 4 months into uni unless you took a gap year so idk
4:02 why is the line on that direction? 3/4Pi =5/4 Pi?
hyun seul -3/4 pi turns clockwise from the initial line
ExamSolutions thank u!
how come u didn't use the distance formula to find the cartesian equation? cuz you're not equating it to another complex no? cuz I did the cartesian equation like in your perpendicular bisector vid and got a different answer
why did you you put the imaginary over the real rather than the real over imaginary at 9:25
+hoptheloop opposite over adjacent
+hoptheloop It is always imaginary over real. Check this out www.examsolutions.net/maths-revision/further-maths/complex-numbers/arg-mod/tutorial-1.php
I don't understand the tan part and I thought it could only be used for right angled triangles.
Try looking back at this www.examsolutions.net/tutorials/modulus-argument-complex-number/?board=Edexcel&module=fp1&topic=1745 I hope it helps answer your question