complex numbers (2 Viewers)

[RohitShubesh21]

Hi for last question goodsirs..

 

[vishnay]

year 9s already starting complex

now that's dedication

 

[CM_Tutor]

Expand the RHS from (b), which is equal to , to get the values for



and



These are the values of and , the roots of the quadratic with the desired roots.

You should get that for a non-zero constant.

 

[RohitShubesh21]

Expand the RHS from (b), which is equal to , to get the values for



and



These are the values of and , the roots of the quadratic with the desired roots.

You should get that for a non-zero constant.

OhOk Last help for this question sir.

 

[icycledough]

You can go about it by converting it into polar form, so you can equate the equation above to cos(theta) + i*sin(theta). Then, getting the right modulus and argument, you can graph the roots on a diagram

 

[RohitShubesh21]

You can go about it by converting it into polar form, so you can equate the equation above to cos(theta) + i*sin(theta). Then, getting the right modulus and argument, you can graph the roots on a diagram

Yes I am done that but I am not getting answer I think my argument is wrong. I have got pi/6 but when I add 2kpi it is getting wrong answer.

 

[icycledough]

Yes I am done that but I am not getting answer I think my argument is wrong. I have got pi/6 but when I add 2kpi it is getting wrong answer.

As the Cartesian equation has 2 negatives, the argument should lie in the 4th quadrant. Thus, your argument should be a negative, acute angle.

 

[CM_Tutor]

As the Cartesian equation has 2 negatives, the argument should lie in the 4th quadrant. Thus, your argument should be a negative, acute angle.

I think you mean the third quadrant...

 

[icycledough]

I think you mean the third quadrant...

Oh yep, my bad. I accidentally referenced the bottom right quadrant as the 1st quadrant instead of the top right

 

[CM_Tutor]

And so, in polar form, the complex number has a negative obtuse angle

 

[Velocifire]

year 9s already starting complex

now that's dedication

yea I thought I was smart when I was the only person who could recite the quadratic formula by heart in year 9

 

[vishnay]

yea I thought I was smart when I was the only person who could recite the quadratic formula by heart in year 9

i thought i was smart when i could multiply fractions in kindy

 

[RohitShubesh21]

Sorry I dont understand yet. Could you do working out so I can see how to do... sorry verymuch

 

[ExtremelyBoredUser]

yea I thought I was smart when I was the only person who could recite the quadratic formula by heart in year 9

Hahah lol same I guess not

 

[idkkdi]

Expand the RHS from (b), which is equal to , to get the values for



and



These are the values of and , the roots of the quadratic with the desired roots.

You should get that for a non-zero constant.

alternatively,

1+z+z^2+z^3+z^4 = 0
(z^2+z^-2) + (z+z^-1) + 1 = 0
2cos2t + 2cost + 1 = 0
4cos^2t + 2cost - 1 = 0
Let cost = x,
hence 4x^2+2x-1 = 0 for Re(z), where z=/1, z is a 5th root of unity, which is cos2pi/5, cos4pi/5.

 

[ExtremelyBoredUser]

Sorry I dont understand yet. Could you do working out so I can see how to do... sorry verymuch

Now you would get the angle to be pi/3 when you do tan^1(-8sqrt(3)/-8) however that does not necessarily mean that the argument is pi/3. Construct a diagram and plot the complex number, you will find it lies on the 3rd quadrant. What you are doing with tan^1(-8sqrt(3)/-8) is you are finding the angle of that triangle specifically in that quadrant, not the actual argument itself. Hence the argument would be -2pi/3 when you do pi - pi/3. I believe this is the error you made - everything else is just routine.

 

[RohitShubesh21]

alternatively,

1+z+z^2+z^3+z^4 = 0
(z^2+z^-2) + (z+z^-1) + 1 = 0
2cos2t + 2cost + 1 = 0
4cos^2t + 2cost - 1 = 0
Let cost = x,
hence 4x^2+2x-1 = 0 for Re(z), where z=/1, z is a 5th root of unity, which is cos2pi/5, cos4pi/5.

Thank you sir,... I am asumming the identity z^n + z^-n = 2cosnt .. is it assumed or I need to prove sir?.

 

[RohitShubesh21]

View attachment 32881

Now you would get the angle to be pi/3 when you do tan^1(-8sqrt(3)/-8) however that does not necessarily mean that the argument is pi/3. Construct a diagram and plot the complex number, you will find it lies on the 3rd quadrant. What you are doing with tan^1(-8sqrt(3)/-8) is you are finding the angle of that triangle specifically in that quadrant, not the actual argument itself. Hence the argument would be -2pi/3 when you do pi - pi/3. I believe this is the error you made - everything else is just routine.

Drawing it up is pretty straightforward, if you do need a diagram feel free to ask. Also its commendable that you have started this topic at Year 9 already, and you have the initiative to do so! Don't feel shy to ask for more help in the forums as I'm sure more people will be willing to help out. Welcome.

Ohok, thanks very muchsir.

 

[idkkdi]

Thank you sir,... I am asumming the identity z^n + z^-n = 2cosnt .. is it assumed or I need to prove sir?.

prove it if you want to play it safe.

 

[CM_Tutor]

Thank you sir,... I am asumming the identity z^n + z^-n = 2cosnt .. is it assumed or I need to prove sir?.

If you were just to assert it, make sure you note that it is only true if |z| = 1, and show that condition is satisfied in the circumstances in which you use it.