Complex Number Question

[Sanguinne]

Not sure if Im suppose to post this general maths advanced but w/e

Find all the solutions of the equation z⁴ - 2z² + 4 = 0 in polar form.

I have no clue how to work this out.

[Alwin]

Not sure if Im suppose to post this general maths advanced but w/e

Find all the solutions of the equation z⁴ - 2z² + 4 = 0 in polar form.

I have no clue how to work this out.

let u = z², solve for u then sub z² back in and solve for z (de moivre's theorem)

[M_BONG]

Not sure if Im suppose to post this general maths advanced but w/e

Find all the solutions of the equation z⁴ - 2z² + 4 = 0 in polar form.

I have no clue how to work this out.

Are the solutions:
root2 cis (-pi/3)   root2 cis (2pi]/3},  root2 cis (-pi6) , root 2 cis (5pi/6)?

[Homer]

something like this was on this years Spesh Exam 1

[LOLs99]

Are the solutions:
root2 cis (-pi/3)   root2 cis (2pi]/3},  root2 cis (-pi6) , root 2 cis (5pi/6)?

Why is it root2?
I think the soln are 2 cis (pi/3), 2cis(2pi/3), 2cis (-2pi/3), 2 cis (-pi/3) .
I just do it in my head now.. Shld be 4 soln from 4 quadrant cuz of the z^2.

[M_BONG]

Why is it root2?
I think the soln are 2 cis (pi/3), 2cis(2pi/3), 2cis (-2pi/3), 2 cis (-pi/3) .
I just do it in my head now.. Shld be 4 soln from 4 quadrant cuz of the z^2.

Um, it is root2.
Just type it into the CAS.

Csolve(...)
I just used one solution and converted into polar form. R = root2.

Might want to check your methods,  don't think you can do this in your head..


But the way to do this question tech free is to:
1. Sub y=z^2 into equation
2. Complete the square.
3. You won't get a different of squares, therefore use "i^2"
4. Solve for y
5. Remember y=z^2.
6. Find the square roots of z^2 answers obtained.
7. Use demoivre's theroem, remembering Arg is between -pi and pi

[LOLs99]

Um, it is root2.
Just type it into the CAS.

Csolve(...)
I just used one solution and converted into polar form. R = root2.

Might want to check your methods,  don't think you can do this in your head..


But the way to do this question tech free is to:
1. Sub y=z^2 into equation
2. Complete the square.
3. You won't get a different of squares, therefore use "i^2"
4. Solve for y
5. Remember y=z^2.
6. Find the square roots of z^2 answers obtained.
7. Use demoivre's theroem, remembering Arg is between -pi and pi

Ahh that should be right. I forget the square root for the soln .  Never do complex stuff in your head haha

[RKTR]

Are the solutions:
root2 cis (-pi/3)   root2 cis (2pi]/3},  root2 cis (-pi6) , root 2 cis (5pi/6)?

Why do I get pi/6 and -5pi/6 for the first two?

[Phy124]

I do it like this:











However I don't know if everyone can convert to polar form as easily.

[M_BONG]

I do it like this:











However I don't know if everyone can convert to polar form as easily.

These answers are definitely right. I accidentally switched the sign around in one of the terms. So I got -1+3i instead of 1+3i.
Disregard my answers please.

[lzxnl]

z^4-2z^2+4 = (z^6 + 8 )/(z^2 + 2)
Solve z^6 + 8 =0 disregarding the imaginary solutions where z^2+2=0

[hyunah]

there is a complex number question:
1 + i, (1 + i)^2, (1 + i)^3, (1 + i)^4
describe the geometrical pattern observed in the position of these complex numbers?

[lzxnl]

Rewrite each term in polar form; does it become more apparent?

[hyunah]

i get that it turns anticlockwise, but i dont get it when they say by pi/4?

[b^3]

So we can see that for each the magnitude is of the previous and we rotate through an angle of (as we have a multiple of each time).

https://www.desmos.com/calculator/ahgopzmhlx
I've drawn the lines in to try and help show the angle but you don't need them when you just have the point.