I've watched this whole series from the beginning and, as of this video, have surpassed all the mathematics I ever learned in high school. I always assumed this stuff would be out-of-reach for the rest of my life, but here I am at the age of 30 looking forward to trig and calculus. Thank you, Professor Dave!
@@truesanatani2892 Like square roots of positive numbers, there are two square roots of -1. By convention, without specifying otherwise, we are interested in the positive imaginary square root of -1, which is +i, and which is CCW from the positive real numbers on the complex plane.
In the last questions i still don't get it why it come up 2+4i in the subtraction questions when i try it it was 2+18i could u please tell me i think i made a mistake but where
I imagine you didn't distribute the negative sign across the second term. (6 - 7i) - (4 - 11i ), as the negative is outside the bracket, you have to distribute the negative sign across both terms in the second term. (6 - 7i) (-4 + 11i), then you can remove brackets and solve. 6 - 7i - 4 + 11i 2 + 4i
X+1=5 How many solutions of this equation? One. that is x=4 X^2+1=5 by using the quadratic formula we get x= 2, -2 X^3-4x^2-x=-6 by using synthetic division we get (x-3)(x-2)(x+1)=0 .so x= -1,2,3 See any pattern?
Complex Analysis by lang (4th edition) : book uses alpha=a+bi.... def: "The real number b is called the imaginary part of alpha, and denoted by Im(alpha)"
well i don't know what that book is or what it's trying to say, but if you have 2 + 3i, the imaginary part is not 3, as 3 is not imaginary. 3i is imaginary, so bi is a pure imaginary number.
There are plenty of applications of imaginary numbers that have nothing to do with faster than the speed of light. For instance, the impedance of a capacitor is -i/(2*pi*f*C), or as electrical engineers like to state it, 1/(j*omega*C), because i is already spoken-for to stand for current.
I've watched this whole series from the beginning and, as of this video, have surpassed all the mathematics I ever learned in high school. I always assumed this stuff would be out-of-reach for the rest of my life, but here I am at the age of 30 looking forward to trig and calculus. Thank you, Professor Dave!
You're a life saver☺️
i got 4 out of 10 in my exam and now today at last i got the concept and your test at the end of the video was amazing i did it correct thankk you
Good timing. Math exam on Monday
Wow sir ur video is that which is really useful for me .I check many videos on utube but I found ur and that was best 🎉😮 THANK YOU SIR
Thank you professor dave!
My broer you're blessing no student or researcher can deny 🙌🏽🔥 thank you once again
Could you please do a video on arguments of complex numbers. It would be very helpful. 🙂
Thanks sana Mwalimu..
Great videos. Do you have one on de Moivre's theorem?
thanks for clear explanation
Your explanation is accurate sir😊
Very helpful information sir ❤❤
It's helpful..thanks
Much love from Tanzania
a casual: i^2 =-1
me, an intelectual: j^2=-1
That's just because i is spoken-for to stand for current, so its alphabet neighbor is used in its place.
Awesome vid! Thanks!
You have made my headache cease. Thank you professor
wow thank you so much
Is the square root of -1 + or - i
positive
Wait... if there was a quartic polynomial with imaginary solutions. i squared is -1. How could that make a real solution?
@@ProfessorDaveExplains how is it positive professor..
I mean you yourself have written -1 there..
i= square root of minus 1
i squared = minus 1..
Yes, he asked if the square root of -1 was plus or minus i. It's positive i. Therefore i squared is -1.
@@truesanatani2892 Like square roots of positive numbers, there are two square roots of -1. By convention, without specifying otherwise, we are interested in the positive imaginary square root of -1, which is +i, and which is CCW from the positive real numbers on the complex plane.
thank you!!
thanks
Wow thanks ❤❤😊
Interesting
thanks 🙏🙏
thank you!! clear explanations;
In the last questions i still don't get it why it come up 2+4i in the subtraction questions when i try it it was 2+18i could u please tell me i think i made a mistake but where
I imagine you didn't distribute the negative sign across the second term.
(6 - 7i) - (4 - 11i ), as the negative is outside the bracket, you have to distribute the negative sign across both terms in the second term.
(6 - 7i) (-4 + 11i), then you can remove brackets and solve.
6 - 7i - 4 + 11i
2 + 4i
At 5:39 how did that 15 changed into 13 ? Is it because of 15 and (-2) ? = 13 😅
15 + (-2)
15 - 2
13
@@pinkfluffyleona thnx i was confused 😂
Please how did you arrive at 20+16i😢
Thank you for the explanation. It really helped!
Going for advance Mathematics
i didn't get why the number of solutions must be the same number of n
X+1=5 How many solutions of this equation? One. that is x=4
X^2+1=5 by using the quadratic formula we get x= 2, -2
X^3-4x^2-x=-6 by using synthetic division we get (x-3)(x-2)(x+1)=0 .so x= -1,2,3
See any pattern?
Is 4th one is correct in comprehension?
The fourth question is correct, or at least, I got the same answer as the solution.
The imaginary part of a+bi is just b
nope! if it was just b, it would be a real number, you need the i for it to be imaginary.
Complex Analysis by lang (4th edition) : book uses alpha=a+bi.... def: "The real number b is called the imaginary part of alpha, and denoted by Im(alpha)"
well i don't know what that book is or what it's trying to say, but if you have 2 + 3i, the imaginary part is not 3, as 3 is not imaginary. 3i is imaginary, so bi is a pure imaginary number.
Complex Variables with Applications by Wunsch: "We say that a is the real part of z and that b is the imaginary part."
yes bi is a pure imaginary number whose imaginary part is b.
How did you arrive at 2+4i😢
Are you stupid bro?
6-4= 2 and -7i - -11i = -7i + 11i = 4i
this was in 12th and 11th now I'm in degree 😅
im imsoniac now
So “complex”!
If you are here because you're writing an exam soon let's gather here
Imaginary number only exist if you travel faster than light
you're thinking of imaginary time/space, due to the radical in the lorentz transformation equations. this is just good old abstract mathematics here!
@@ProfessorDaveExplains wow
There are plenty of applications of imaginary numbers that have nothing to do with faster than the speed of light. For instance, the impedance of a capacitor is -i/(2*pi*f*C), or as electrical engineers like to state it, 1/(j*omega*C), because i is already spoken-for to stand for current.
Wtf is this 2010 intro
thanks a lot, I'm improving i