Answer:
z = 2, -1 + sqrt(3)i, -1 + sqrt(3)i
Step-by-step explanation:
z³ = 8[cos(0) + isin(0)]
z = 8^⅓ = 2
z = {8[cos(2pi) + isin(2pi)]}^⅓
z = 2[cos(2pi/3) + isin(2pi/3)]
z = -1 + sqrt(3)i
z = {8[cos(4pi) + isin(4pi)]}^⅓
z = 2[cos(4pi/3) + isin(4pi/3)]
z = -1 - sqrt(3)i
The cube roots of 8 in the form of a + ib is −1 ± √3i after solving with concept of complex number.
It is defined as the number which can be written as x+iy where x is the real number or real part of the complex number and y is the imaginary part of the complex number and i is the iota which is nothing but a square root of -1.
Let's suppose the cube root of 8 is a + ib
∛8 = a + ib
cubing on the both side:
8 = (a + ib)³
After expanding the (a + ib)³
8 = (a³ - 3ab²) + i(3a²b - b³)
Equating real and imaginary part:
a³ - 3ab² = 8 and
3a²b - b³ = 0
3a²b = b³
3a² = b²
Plug the value in the other equation;
a³ - 3a(3a²) = 8
a³ - 9a³ = 8
-8a³ = 8
a³ = -1
[tex]\rm a=-1,\:x=\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2},\:x=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}[/tex]
Similarly, we can find the value of b by plugging a in the equation.
The cube roots of 8 in the form of a + ib is −1 ± √3i
Thus, the cube roots of 8 in the form of a + ib is −1 ± √3i after solving with concept of complex number.
Learn more about the complex number here:
brainly.com/question/10251853
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