Find the principle root (-64)1/6. Express the result in the form a+bi.
(2i)^6= -64 = (-64)1/6= 0+2i
I got it wrong, please help.
you have 64@180 on the complex plane.
the sixth root of that is ..
2@30
in rectangular coordinates...
2cos30+2isin30
Try that.
Ok but are you sure it's ok for how they want me to express it?
To find the principal root of (-64)^(1/6) and express it in the form a + bi, you need to follow a step-by-step process.
Step 1: Rewrite the expression in exponential form.
(-64)^(1/6) can be written as [(-64)^(1/6)]^6. Since raising a power to a power allows us to cancel the exponents, this simplifies to (-64)^((1/6)*6) = (-64)^1 = -64.
Step 2: Identify the principal root of -64.
The principal root of -64 is the number whose sixth power equals -64. In this case, (-2)^6 = 64, so the principal root must be -2.
Step 3: Express the result in the form a + bi.
The final step is to express the principal root in the form a + bi, where a and b are real numbers. In this case, since an imaginary number is involved, we can rewrite as -2 + 0i.
Therefore, the principal root of (-64)^(1/6) expressed in the form a + bi is -2 + 0i.