Find the Taylor series expansion of(a) f (z) = 6z + 8(2z + 3)(4z + 5) at z = 1, (b) f (z) = 2i4 + iz at z = −3i

Sure, let's solve this step by step.

(a) f (z) = 6z + 8(2z + 3)(4z + 5) at z = 1

First, simplify the function: f(z) = 6z + 8(2z + 3)(4z + 5) = 6z + 8(8z^2 + 22z + 15) = 6z + 64z^2 + 176z + 120 = 64z^2 + 182z + 120

The Taylor series expansion of a function about a point can be found using the formula:

f(a) + f'(a)(z-a) + f''(a)(z-a)^2/2! + f'''(a)(z-a)^3/3! + ...

Here, a = 1. So, we need to find the derivatives of the function at z = 1.

f'(z) = 128z + 182 f''(z) = 128

At z = 1,

f(1) = 641^2 + 1821 + 120 = 366 f'(1) = 128*1 + 182 = 310 f''(1) = 128

So, the Taylor series expansion of the function at z = 1 is:

f(z) = 366 + 310(z-1) + 128(z-1)^2/2! + 0(z-1)^3/3! + ...

Simplifying this, we get:

f(z) = 366 + 310z - 310 + 64z^2 - 128z + 64 = 64z^2 + 182z + 120

which is the original function, as expected.

(b) f (z) = 2i4 + iz at z = −3i

This function is a linear function, and its Taylor series expansion is the function itself. So, f(z) = 2i4 + iz for all z.