We are given complex numbers $z_1 = 3 - i$, $z_2 = 2 - 3i$, and $z_3 = -6 + i$, with a relation for $Z$ expressed as:

$Z = z_2 + \frac{z_1 + z_3}{z_1 \cdot z_3}$

We need to find $E = I \cdot Z$ when $I = 2 - 4i$.

Step 1: Find $z_1 + z_3$

$z_1 + z_3 = (3 - i) + (-6 + i) = -3$

Step 2: Find $z_1 \cdot z_3$

$z_1 \cdot z_3 = (3 - i)(-6 + i) = 3(-6 + i) - i(-6 + i)$ $= -18 + 3i + 6i + 1 = -17 + 9i$

Step 3: Calculate $\frac{z_1 + z_3}{z_1 \cdot z_3}$

$\frac{-3}{-17 + 9i}$

To simplify, multiply the numerator and denominator by the conjugate of the denominator: $\frac{-3}{-17 + 9i} \cdot \frac{-17 - 9i}{-17 - 9i} = \frac{-3(-17 - 9i)}{(-17)^2 + (9)^2}$ $= \frac{51 + 27i}{289 + 81} = \frac{51 + 27i}{370}$ $= \frac{51}{370} + \frac{27i}{370}$

Step 4: Calculate $Z$

$Z = z_2 + \frac{z_1 + z_3}{z_1 \cdot z_3}$ $Z = (2 - 3i) + \left( \frac{51}{370} + \frac{27i}{370} \right)$ $= \left( 2 + \frac{51}{370} \right) + \left( -3i + \frac{27i}{370} \right)$ $= \frac{740 + 51}{370} + i\left( \frac{-1110 + 27}{370} \right)$ $= \frac{791}{370} + i\left( \frac{-1083}{370} \right)$ $Z = 2.138 + i(-2.927)$

Step 5: Find $E = I \cdot Z$

We know $I = 2 - 4i$, so: $E = (2 - 4i)(2.138 - 2.927i)$ Using distribution: $E = 2(2.138 - 2.927i) - 4i(2.138 - 2.927i)$ $E = 4.276 - 5.854i - 4i(2.138 - 2.927i)$ $= 4.276 - 5.854i - 8.552i + 11.708$ $E = (4.276 + 11.708) + (-5.854 - 8.552)i$ $E = 15.984 - 14.406i$

Thus, $E = 15.984 - 14.406i$.

Would you like further details on any step or additional explanation?

Here are 5 related questions for practice:

1. Calculate $z_1 \cdot z_2$ and simplify.
2. Find $|Z|$ (the modulus of $Z$).
3. What is the argument (angle) of $I$ in polar form?
4. Express $Z$ in polar form.
5. Calculate $E$ if $I = 1 + i$.

Tip: When multiplying complex numbers, remember to use the distributive property carefully and combine like terms (real and imaginary parts).