1. Find (-1+i)^1/3
2. Express 2(cos5pi/3 + isin5pi/3) in regular form.
would this be correct: 1-√3i?
3. Find (2+4i)^3 in regular form.
4. Express 5(cos5pi/6 + isin5pi/6) in regular form.
would this be correct?: -5√3/2 +5/2i?
I find on powers of complex numbers, it is easier to just go to polar form.
For instance
(-1+i)^(1/3)
put it in polar..
(sqrt2 @135 )^1/3
2^(1/6)@135/3= 1.122@45
= 1.122cos45+i*1.122sin45
= 1.122*.707 + i*1.122*.707
then the other one
(2+4i)=sqrt20@[email protected] so
(2+4i)^3= 20^1.5 @87.14*3=89.4@267
= 89.4(cos267+isin267) and you can simplify that.
#2 and #4 are correct
1. To find the cube root of a complex number, we can use the polar form and De Moivre's formula.
First, let's write (-1 + i) in polar form. The magnitude (r) can be found using the formula √(a^2 + b^2), where a = -1 and b = 1. So, r = √((-1)^2 + 1^2) = √2.
The argument (θ) can be found using the formula θ = arctan(b/a), where a = -1 and b = 1. So, θ = arctan(1/-1) = -π/4.
Now, to find the cube root, we use De Moivre's formula:
(z^n)^(1/n) = z^(1/n).
Let's plug the values in:
(-1 + i)^(1/3) = (r * cis(θ))^(1/3)
= √2 * cis(-π/12)
To convert this back to rectangular form, we can use Euler's formula:
cis(θ) = cos(θ) + i * sin(θ).
Multiplying √2 by cos(-π/12) and sin(-π/12), we get:
-√(2/2) + i * √(2/2)
= -√2/√2 + i * √2/√2
= -1 + i
So, the cube root of (-1 + i) is -1 + i.
2. To express 2(cos(5π/3) + i*sin(5π/3)) in regular form, we can use Euler's formula.
Euler's formula states that e^(ix) = cos(x) + i*sin(x).
So, in this case, we have:
2 * (cos(5π/3) + i*sin(5π/3))
= 2 * e^(5πi/3)
Now, let's rewrite 2 as 2*e^(0i):
2*e^(0i) * e^(5πi/3)
= 2 * e^(5πi/3) * e^(0i)
= 2 * e^(5πi/3 + 0i)
= 2 * e^(5πi/3)
= 2 * cos(5π/3) + 2i * sin(5π/3)
Evaluating cos(5π/3) and sin(5π/3):
2 * (-1/2) + 2i * (√3/2)
= -1 + i√3
So, 2(cos(5π/3) + i*sin(5π/3)) is -1 + i√3.
3. To find (2 + 4i)^3 in regular form, we can simply expand it using the binomial theorem or by multiplying it out.
(2 + 4i)^3 = (2 + 4i)(2 + 4i)(2 + 4i)
= (2 * 2 + 2 * 4i + 4i * 2 + 4i * 4i)(2 + 4i)
= (4 + 8i + 8i - 16)(2 + 4i)
= (-12 + 16i)(2 + 4i)
= -24 + 8i + 32i + 64i^2
= -24 + 40i + 64(-1)
= -24 + 40i - 64
= -88 + 40i
So, (2 + 4i)^3 in regular form is -88 + 40i.
4. To express 5(cos(5π/6) + i*sin(5π/6)) in regular form, we can use Euler's formula.
Using Euler's formula, we have:
5 * (cos(5π/6) + i*sin(5π/6))
= 5 * e^(5πi/6)
Expanding e^(5πi/6):
e^(5πi/6)
= cos(5π/6) + i*sin(5π/6)
= -√3/2 + i*(1/2)
Multiplying this by 5:
5 * (-√3/2 + i*(1/2))
= -5√3/2 + 5i/2
So, 5(cos(5π/6) + i*sin(5π/6)) in regular form is -5√3/2 + 5i/2.