Question

This question deals with complex numbers.

(a) Find the cubic roots of √5 + i√13, namely all the complex numbers z such that z3 = √5 + i√13

Express the roots in polar form.

Answer 1

To find the cubic roots of √5 + i√13 and express them in polar form, we can convert the given number to polar form by calculating the magnitude and the argument. The roots in polar form are (√5 + √13, atan(√13/√5)), (√5 + √13, atan(√13/√5) + 2π/3), and (√5 + √13, atan(√13/√5) + 4π/3).

Step-by-step explanation:

To find the cubic roots of √5 + i√13 and express them in polar form, we can start by expressing the given number in rectangular form. √5 + i√13 can be written as (0 + √5) + (1√13)i.

Now, we can express the number in polar form by calculating the magnitude and the argument. The magnitude of the number is the square root of the sum of the squares of the real and imaginary parts, which is √(0^2 + √5^2 + (1√13)^2) = √5 + √13. The argument can be found using the inverse tangent function: arg = atan(1√13 / 0 + √5) = atan(√13 / √5) = atan(√13 / √5) = atan(√13/√5) = atan(√#√5).

Therefore, the roots in polar form are:

Root 1: (√5 + √13, atan(√13/√5))

Root 2: (√5 + √13, atan(√13/√5) + 2π/3)

Root 3: (√5 + √13, atan(√13/√5) + 4π/3)