Roots of complex numbers (1 Viewer)

[Grey Council]

expand the brackets:
z^4=8(SQRT3+i)
= 8sqrt3 + 8i
sqrt(64+64*3) = r = 16 (according to keypad, just check it)
theta = arc tan (1/sqrt3)
theta = pi/3 (is it pi/6? bah, whatever, you get the point)

and work from there

 

[mazza_728]

I still dont get it where did 64 come from??
Wheres ND??
Ok
Let z=r(cos@+isin@)=rcis@
8(sqrt3+i)=8sqrt3+8i
z⁴=r⁴cis4@=8sqrt3+8i
Where do i go from here??
i think my mind has shut down.

 

[Constip8edSkunk]

z^4 = 16cis[ pi/6 ]
z = 2cis[ (12k+1)pi / 24 ]

solutions when k= 0,1,2,3,

 

[mazza_728]

constip8ed skunk did u just halve the angles and double r?

 

[ND]

Originally posted by mazza_728

Wheres ND??

I'm overseas. (i probably won't be around for a while)

Let z=r(cos@+isin@)=rcis@
8(sqrt3+i)=8sqrt3+8i
z4=r4cis4@=8sqrt3+8i
Where do i go from here??
i think my mind has shut down.

All that you have to do(as constip8edskunk did), is to get it into mod-arg form. Remember that:

a+ib=rcis@
where r=sqrt(a^2+b^2) @=arctan(b/a). (note: arctan means inverse tan)

sqrt(64+64*3) = r = 16 (according to keypad, just check it)

It doesn't take long to check it: sqrt(64+3*64)=sqrt(4*64)=2*8=16.

 

[Constip8edSkunk]

yep, 8[sqrt3+i]=8[2cis(pi/6)]=16cis(pi/6)

 

[mazza_728]

Gotcha thankyou oxox

 

[mazza_728]

Ive got another question .. sorry


Find the fifth roots of 32

z⁴=32
Let z=rcis@
32=32(1+0i)=32cis2pi
z⁵=r⁵cis5@
r=2
5@=2pi+2kpi
@=2pi/5+2kpi/5
k=0 z=2cis2pi/5
k=1 z=2cis4pi/5
k=-1 z=2
k=2 z=2cis6pi/5
k=-2 z=2cis-2pi/5

When they ask for the fifth roots are u just looking for 5 roots?

The answer however is 2, 2cis(+/- {2pi/5}), 2cis (+/- {4pi/5})

Did i do something wrong? I think im just not understanding the method properly.

Thankyou

 

[J0n]

You got it right, it is just that 2cis6pi/5 is the same as 2cis-4pi/5.

 

[mazza_728]

thankyou .. i used 2pi instead of 0 which would give u the same degrees etc.

I have another question.. i keep getting on rolls and then stopping and i feel soo stupid to keep posting questions but i cant do this:

Find the fourth roots of 8? is there a trick ??

 

[J0n]

Wouldn't it be pretty much the same as the previous question? I don't see any trick.

 

[mazza_728]

but you'd have to work with the 4th root of 8 which would be hard.. is there another way?

 

[J0n]

z⁴ = 8
|z|⁴cis 4@ = 8cis 0
8^1/4cis 4@
@ = 2kpi/4
k = 1 = pi/4
k = -1 = -pi/4
k = 2 = pi
k = 0 = 0
Roots are 8^1/4cis +/- (pi/4) and 8^1/4cis pi and 8^1/4cis 0

 

[J0n]

Calculator?

 

[mazza_728]

sarcasm noted..
thankyou

 

[mazza_728]

OMG IM SO STUPID!!!

three points of which 1+sqrt3i is one, lie on the circumference of a circle of radius 2 unit and centre at the origin. if these three points form the vertices of an equilateral triangle, find the other two points...

also i dont understand anything about subsets of the complex plane i have a huge problem in interpreting questions! anyone got any tips or advice?

 

[J0n]

To interpret questions better, you can represent the question mathematically. e.g In the above question, there are three points - let them be z₁, z₂, z₃
Next, it says that the points lie on a circle of circumference 2 with radius at .'. the moduli are two.
After that it says that the points for make an equilateral triangle - that means that their principal arguments differ by 2pi/3.

z₁ = 1 + sqrt(3)i = 2cis(pi/3)
.'. The other two points are:
z₂ = 2cis[(pi/3) + (2pi/3)] = 2cis(pi)
z₃ = 2cis[(pi/3) - (2pi/3)] =2cis(-pi/3)

 

[Grey Council]

hrm, how bout this.

Complex roots occur in complex cunjugates. You prolly know this, just haven't seen/thought about consequences. Anyway, what that means is that the roots of any eqation is going to form the vertices of a regular polygon.

ANYWAY, back to the question.

three points of which 1+sqrt3i is one, lie on the circumference of a circle of radius 2 unit and centre at the origin. if these three points form the vertices of an equilateral triangle, find the other two points...

equilateral triangle. Therefore, find the angle that 1+sqrt3i makes with the origin, then rotate it 120 degrees (Because isn't that the angle that a point from the corner of an equilaterla traingle to the centre of that traingle makes?) to the left and then to the right to find the other two points.

You can try and do this graphically, without algebra (if you don't like bringing r CIS theta and i and complex numbers in general) or you can use Jon's method:

z1 = 1 + sqrt(3)i = 2cis(pi/3)
.'. The other two points are:
z2 = 2cis[(pi/3) + (2pi/3)] = 2cis(pi)
z3 = 2cis[(pi/3) - (2pi/3)] =2cis(-pi/3)