Complex Numbers - Loci : Half-lines : ExamSolutions Maths Video Tutorials

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Tutorial on loci and half-lines when dealing with complex numbers

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This was so much easier to follow than my lecturer. Thank you sir

inhopeofabettername
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Omg my teacher used your video to give us notes. This is exactly what I have in my exercise book

memyself
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Thx sir and can you explain demorgan's law also

whitedeviltechvijay
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At 9:13 how d'you get to using tan?

ArtKickers
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why did you you put the imaginary over the real rather than the real over imaginary at 9:25

hoptheloop
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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

mrpeterkempner
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At 9:25 Im confused on how the tan of the obtuse angle equals opp over adjacent

a_dang
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i had a doubt, but you cleared it in this video. I can't thank you enough.

Crazyfingers
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9:45 i get that you using tan, but are you not applying it wrong here? Isn’t arg(z) = -pi + tan(1/4pi) ? Since the angle tan is the one which is inside the triangle and ≠ theta ? And therefore the modulus is completely skewed now as it is not = tan(-3pi/4 )

ramsay
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I don't understand the tan part and I thought it could only be used for right angled triangles.

marquez
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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

pyschoticwaif
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Subscribed, explained much better than my teacher

tlhe_ovIyloS
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This is wonderful thank you :) I was just wondering why x < 3 and not x<= 3 as it looks like the line starts at (3, -2) so x at the start does have the value of 3? Am I misunderstanding something? Thank you :D

eliseratcliffe
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4:02 why is the line on that direction? 3/4Pi =5/4 Pi?

jessecruz
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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

jessecruz