Here are the solutions. 1. Express z = (40^) - i(40^) in a true polar form. 2. Find the imaginary part of (1)/(1+i). 3. Find (x,y); x,y R, if 3y+2xi = x+i+10. 4. Evaluate the following powers of i: a)

Here are the solutions.

1. Express $z = \cos(40^\circ) - i\sin(40^\circ)$ in a true polar form.

The standard polar form is $z = r(\cos\theta + i\sin\theta)$.
Given $z = \cos(40^\circ) - i\sin(40^\circ)$.
Using $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$:

$$z = \cos(-40^\circ) + i\sin(-40^\circ)$$

The modulus $r=1$. The argument $\theta = -40^\circ$.
The final answer is $\boxed{z = \cos(-40^\circ) + i\sin(-40^\circ)}$.

2. Find the imaginary part of $\frac{1}{1+i}$.

Multiply by the conjugate of the denominator:

$$\frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i}$$ $$= \frac{1-i}{1^2 - i^2}$$ $$= \frac{1-i}{1 - (-1)}$$ $$= \frac{1-i}{2}$$ $$= \frac{1}{2} - \frac{1}{2}i$$

The imaginary part is the coefficient of $i$.
The final answer is $\boxed{-\frac{1}{2}}$.

3. Find $(x,y)$; $x,y \in \mathbb{R}$, if $3y+2xi = x+i+10$.

Equate the real parts and the imaginary parts:
Real parts: $3y = x+10 \quad (1)$
Imaginary parts: $2x = 1 \quad (2)$

From (2):

$$x = \frac{1}{2}$$

Substitute $x$ into (1):

$$3y = \frac{1}{2} + 10$$ $$3y = \frac{1}{2} + \frac{20}{2}$$ $$3y = \frac{21}{2}$$ $$y = \frac{21}{2 \times 3}$$ $$y = \frac{7}{2}$$

The final answer is $\boxed{\left(\frac{1}{2}, \frac{7}{2}\right)}$.

4. Evaluate the following powers of $i$:

Recall that $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$. The powers repeat every 4.

*a) $i^8$:

$$i^8 = (i^4)^2 = 1^2 = 1$$

*b) $i^{11}$:

$$i^{11} = i^{4 \times 2 + 3} = (i^4)^2 \cdot i^3 = 1^2 \cdot (-i) = -i$$

*c) $i^{42}$:

$$i^{42} = i^{4 \times 10 + 2} = (i^4)^{10} \cdot i^2 = 1^{10} \cdot (-1) = -1$$

*d) $i^{105}$:

$$i^{105} = i^{4 \times 26 + 1} = (i^4)^{26} \cdot i^1 = 1^{26} \cdot i = i$$

The final answers are: a) $\boxed{1}$, b) $\boxed{-i}$, c) $\boxed{-1}$, d) $\boxed{i}$.

5. Find the reciprocal of the number $z=i$.

The reciprocal is $\frac{1}{z}$:

$$\frac{1}{i} = \frac{1}{i} \times \frac{-i}{-i}$$ $$= \frac{-i}{-i^2}$$ $$= \frac{-i}{-(-1)}$$ $$= \frac{-i}{1}$$ $$= -i$$

The final answer is $\boxed{-i}$.