Roots of complex numbers (1 Viewer)

[Dumsum]

Q: By expressing -2 - 2i in mod/arg form, and using De Moivre's Theorem, find the cube roots of -2 - 2i.

So far I have:

-2 - 2i = sqrt(8) cis (-3pi/4)

cubert(-2 - 2i) = [sqrt(8) cis (-3pi/4)]^(1/3)
= sqrt(2) cis (-pi/4)

Now I assume there's more than one answer. Is this where all the roots lie equally spaced around the argand diagram or is that just for the roots of unity? Or have I missed something else out already?

 

[KFunk]

Let the cube roots of -2 - 2i = z such that z³ = √8cis(-3π/4) and z = rcisθ

then

z³ = r³cis(3θ ) = √8cis(-3π/4)

hence r = cuberoot(√8) = √2

3θ= (-3π/4) +2kπ where k = 0, ±1, ±2 etc...

therefore θ = -3π/12 + 2kπ/3

Using the above will generate your various values of z. I'll leave it to you to work out which ones are unique .

 

[Slidey]

Q: By expressing -2 - 2i in mod/arg form, and using De Moivre's Theorem, find the cube roots of -2 - 2i.

So far I have:

-2 - 2i = sqrt(8) cis (-3pi/4)

cubert(-2 - 2i) = [sqrt(8) cis (-3pi/4)]^(1/3)
= sqrt(2) cis (-pi/4)

Now I assume there's more than one answer. Is this where all the roots lie equally spaced around the argand diagram or is that just for the roots of unity? Or have I missed something else out already?

KFunk is correct. I'm doing this for practice.

Roots of numbers is an entire sub-topic.
-2-2i=8^1/2.cis(-3pi/4)
z^3=-2-2i=8^1/2.cis(-3pi/4)
.'. z=8^1/6.cis([-3pi/4 + 2k.pi]/3)
z=sqrt2.cis(2kpi/3 - pi/4)
Let k=0, z₁=sqrt2.cis(-45)=sqrt2(1/sqrt2 - i/sqrt2)=1-i
Let k=1, z₂=sqrt2.cis(75)
Let k=2, z₃=sqrt2.cis(195)

To figure those out, note that:
75=30+45
cos75=cos30.cos45-sin30.sin45
=sqrt3/2 * 1/sqrt2 - 1/2.1/sqrt2=(sqrt3-1)/2sqrt2
sin75=sin30cos45+sin45cos30
=(1+sqrt3)/2sqrt2
195=225-30
cos195=cos225cos30+sin225sin30
=-(sqrt3+1)/2sqrt2
sin195=sin225cos30-sin30cos225
=-(sqrt3-1)/2sqrt2

.'.
z₁=sqrt2.cis(-45)=1-i
z₂=sqrt2.cis(75)=(sqrt3-1)/2 + (sqrt3+1)i/2
z₃=sqrt2.cis(195)=-(sqrt3-1)/2 - (sqrt3+1)i/2

 

[Trev]

Practice? It would be easier writing it rather than typing, Slide.
PS. The solution was done by KFunk, not Rench.

 

[~ ReNcH ~]

lol.
I was going to say....I didn't write anything

Credit to KFunk for the solution...

 

[FinalFantasy]

hahaha @ captain saying rench!

 

[KFunk]

To be fair our names and avatars are quite alike.

 

[Slidey]

Pardon me!

 

[~ ReNcH ~]

To be fair our names and avatars are quite alike.

The similarities stand out a mile away.

 

[FinalFantasy]

The similarities stand out a mile away.

hahaha i agree