simple complex number question (1 Viewer)

[joesmith1975]

forgot how to do z^3=8i. Express in x+iy form

 

[sasquatch]

z³=8i

|z| = 8
Arg(z) = tan^-1(8/0)
= pi/2

r³cis3@ = 8cis(pi/2)

r³ = 8
r = 2

3@ = pi/2 + 2kpi
3@ = (pi + 4kpi) / 2
@ = (pi + 4kpi) / 6

When k = 0, @ = pi/6
When k = 1, @ = 5pi/6
When k = -1, @ = -pi/2

hence,
z₁ = 2cis(pi/6)
z₂ = 2cis(5pi/6)
z₃ = 2cis(-pi/2)

 

[joesmith1975]

find all over complex numbers

 

[sasquatch]

I dont really understand what that means..but taking a logical guess...

z₁ = 2cis(pi/6)
=2(cos(pi/6) + isin(pi/6))
=2(root3/2 + i/2)
= root3 + i

z₂ = 2cis(5pi/6)
=2(cos(5pi/6) + isin(5pi/6))
=2(-root3/2 + i/2)
= - root3 + i

z₃ = 2cis(-pi/2)
=2(cos(-pi/2) + isin(-pi/2))
=2(0 - i)
= -2i

 

[XcarvengerX]

z³=8i

|z| = 8
Arg(z) = tan^-1(8/0)
= pi/2

r³cis3@ = 8cis(pi/2)

r³ = 8
r = 2

3@ = pi/2 + 2kpi
3@ = (pi + 4kpi) / 2
@ = (pi + 4kpi) / 6

When k = 0, @ = pi/6
When k = 1, @ = 5pi/6
When k = -1, @ = -pi/2

hence,
z¹ = 2cis(pi/6)
z² = 2cis(5pi/6)
z³ = 2cis(-pi/2)

Then to express it in x + iy form, just do it with your calculator:
z₁ = 2 ( √3/2 + 1/2i )
= √3 + i
z₂ = 2 ( -√3/2 + 1/2i )
= -√3 + i
z₃ = 2 (0 + (-1)i )
= 0 - 2i

 

[sasquatch]

hehe same stuff up as me...did everything to superscript instead of subscript BUT i fixed it...

 

[mr_brightside]

haha <3 that...

"simple complex"

 

[XcarvengerX]

Just edit, and I got the square root sign from Microsoft Word

 

[sasquatch]

Yeah i did edit. ANd you dont need to use Microsoft Word, you can use the character map located at Start > Programs > Accessories > System Tools. I just cant be bothered...

 

[XcarvengerX]

Yeah i did edit. ANd you dont need to use Microsoft Word, you can use the character map located at Start > Programs > Accessories > System Tools. I just cant be bothered...

But I customised my windows so much I can't find it there anymore. And Word is easier I think.

 

[sasquatch]

Oh well whatever works best for you then!