Change it back to a+bi form using m=0,1,2 .Jamin2112 said:Homework Statement
Find all the values of (-2+2i)1/3
Homework Equations
eib = cos(b) + i * sin(b)
The Attempt at a Solution
(-2+2i)1/3 = (2(-1+i))1/3 = 21/3 (√2)1/3 (eiπ(3/4 + 2m))1/3 = 21/2 (eπi)1/4 + 2m/3 = ... Where am I going with this? I don't even know.
To find all the values of (-2+2i)^(1/3), you can use the polar form of complex numbers and the properties of exponents. First, write -2+2i in polar form as 2√2(cos(135°) + isin(135°)). Then, use the property that (r(cosθ+isinθ))^n = r^n(cos(nθ) + isin(nθ)). In this case, n=1/3, so we get (2√2)^(1/3)(cos(135°/3)+isin(135°/3)). This simplifies to 2(cos(45°)+isin(45°)), which is equivalent to 2(√2/2 + i√2/2). Therefore, the three values of (-2+2i)^(1/3) are 2(√2/2 + i√2/2), 2(√2/2(cos(120°)+isin(120°))), and 2(√2/2(cos(240°)+isin(240°))).
Yes, (-2+2i)^(1/3) can have multiple values. This is because complex numbers have multiple roots, and in this case, the cube root of (-2+2i) has three values.
The principal value of (-2+2i)^(1/3) is the value with the smallest positive argument. In this case, it would be 2(√2/2 + i√2/2) or 2(cos(45°)+isin(45°)).
The values of (-2+2i)^(1/3) can be represented as three points on a complex plane. The first point is at 2(√2/2 + i√2/2) or 2(cos(45°)+isin(45°)). The second point is at 2(√2/2(cos(120°)+isin(120°))), and the third point is at 2(√2/2(cos(240°)+isin(240°))). These three points form an equilateral triangle with side length 2√2 on the complex plane.
Finding all the values of (-2+2i)^(1/3) is useful in solving complex equations and understanding the behavior of complex numbers. It also helps in visualizing the geometric properties of complex numbers on a complex plane.