Complex No. (1 Viewer)

[nightweaver066]

If tanR is positive (finding principal argument), then its going to be in the 3rd quadrant and you're going to be working with negative radians -pi < x < -pi/2.

If tanR is negative, then you have to look at which quadrant the complex graph lies on the Argand diagram to determine which quadrant out of the first, second and fourth quadrant it's in.

What i like to do is that whenever i find an angle, i always convert it to its acute form and then determine whether i need to take it away from 180 or add it to something.

 

[Aysce]

Ahh okay thanks I GET IT AT LAST !

 

[Drongoski]

Ahh okay thanks I GET IT AT LAST !

It's a lot easier than that.

Because the principal argument needs to satisfy:
whenever the complex number is in 3rd or 4th quadrant, you figure out the argument (angle) backwards, i,e, clockwise, from the positive x-axis,. For z = -1-i, you get 3pi/4, and when figuring out angle clockwise, the angle is negative: so it's -3pi/4. You don't need to add to or subtract from pi or 180 deg.

 

[Omnipotence]

Picture where would lie on the Argand diagram. -1 would mean you go to the left of the Origin one unit, and means you go below the origin by one unit. This puts the coordinate in the 3rd quadrant, and since you go from you get the Argument as being , which can be simplified.

Nope. Its .

 

[AAEldar]

Nope. Its .

I'm only human.

 

[bleakarcher]

ayye guyz, is complex number hard?

 

[Omnipotence]

Nope, once you do a few questions - it becomes really repetitive.

 

[Hermes1]

ur first test, complex numbers is going to feel slightly uneasy. but when u get into a bit more, it becomes shit easy. in the end, all it is is question 2 of the hsc exam. u shuld be aiming to ace the complex section of ur trials and hsc.

 

[bleakarcher]

wat would u say is harder volumez or complex number? for me, volumez were at first rather intimidating, im not sure about complex numberz

 

[Omnipotence]

Volumes requires a perfect sketch and the solution should really flow out from there. I have/had problems with Conics.

 

[Hermes1]

wat would u say is harder volumez or complex number? for me, volumez were at first rather intimidating, im not sure about complex numberz

volumes can be hard if they want to make it hard. but they usually dont. so u always see volumes round bout Q3-4 and one slightly difficult one at q6.
i reckon they r both relatively easy.

 

[Hermes1]

Volumes requires a perfect sketch and the solution should really flow out from there. I have/had problems with Conics.

conics can be a bitch.

 

[bleakarcher]

ah ok, thanks hermes. wat about integration?

 

[Omnipotence]

Easy. Extended from 3U intergration. (Q1s of HSC and Trial)

 

[Hermes1]

ah ok, thanks hermes. wat about integration?

Integration is the topic u want to get full marks for in ur paper. i mean the only way u could find it difficult is if u get a question like i had in my 4U trial (ill PM it to u when i get it bak) which was a reduction formula question.

also be wary of little tricks like reverse chain rule. also have an attempt at this question:



using integration by parts.

 

[Shadowdude]

Integration is the topic u want to get full marks for in ur paper. i mean the only way u could find it difficult is if u get a question like i had in my 4U trial (ill PM it to u when i get it bak) which was a reduction formula question.

also be wary of little tricks like reverse chain rule. also have an attempt at this question:



using integration by parts.

Set x = a tan(u), right?

 

[Hermes1]

Set x = a sec(u), right?

no it says use integration by parts.

 

[Shadowdude]

If you set x = a tan(u) - not sec(u), my bad! You fudge it until you get a sec^3 integral I believe, and that's when you use parts.

 

[Hermes1]

If you set x = a tan(u) - not sec(u), my bad! You fudge it until you get a sec^3 integral I believe, and that's when you use parts.

you cant do that. u hav to use parts from the outset.

 

[bleakarcher]

(1/2)[xsqrt[a^2+x^2]+a^2*log[e][x+sqrt[x^2+a^2]]]+C
by integration by parts or trigonometric substitution, that question is easy lol