Express the following in a true polar form, z = (40^) - i(40^). 2. Find the imaginary part of (1)/(1+i).

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

• Step 1: Identify the modulus and argument. The given complex number is $z = \cos(40^\circ) - i\sin(40^\circ)$. The modulus $r$ is 1. The standard polar form is $z = r(\cos\theta + i\sin\theta)$. We use the trigonometric identities $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$.

• Step 2: Rewrite the expression in standard polar form.

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

  The angle $-40^\circ$ is a valid argument. Alternatively, we can express the angle as a positive value by adding $360^\circ$: $-40^\circ + 360^\circ = 320^\circ$.

  $$z = \cos(320^\circ) + i\sin(320^\circ)$$

  Both forms are correct. We will use the form with the negative angle. The true polar form is $\boxed{\cos(-40^\circ) + i\sin(-40^\circ)}$.

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

• Step 1: Rationalize the denominator. To find the imaginary part, we need to express the complex number in the form $a+bi$. We multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $1+i$ is $1-i$.

  $$\frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i}$$

• Step 2: Perform the multiplication.

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

• Step 3: Separate the real and imaginary parts.

  $$\frac{1}{2} - \frac{1}{2}i$$

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

3. Find $(x,y)$; $x,y \in \mathbb{R}$, if $3y + 2xi = x+1+10i$.

• Step 1: Equate the real parts. For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. The real part of the left side is $3y$. The real part of the right side is $x+1$.

  $$3y = x+1 \quad (*)$$

• Step 2: Equate the imaginary parts. The imaginary part of the left side is $2x$. The imaginary part of the right side is 10.

  $$2x = 10$$

• Step 3: Solve the system of equations. From $2x=10$, we find $x$:

  $$x = \frac{10}{2}$$ $$x = 5$$

  Substitute $x=5$ into equation $(*)$:

  $$3y = 5+1$$ $$3y = 6$$ $$y = \frac{6}{3}$$ $$y = 2$$

  The values are $x=5$ and $y=2$. So, $(x,y) = \boxed{(5,2)}$.

4. Evaluate the following powers of $i$:

Recall the cycle of powers of $i$:
$i^1 = i$
$i^2 = -1$
$i^3 = -i$
$i^4 = 1$
For any integer $n$, $i^n = i^{n \pmod 4}$. If $n \pmod 4 = 0$, then $i^n = i^4 = 1$.

• a) $i^8$ Divide the exponent by 4: $8 \div 4 = 2$ with a remainder of 0.

  $$i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1$$

  Alternatively, $i^8 = i^{8 \pmod 4} = i^0 = 1$. The value is $\boxed{1}$.

• b) $i^{11}$ Divide the exponent by 4: $11 \div 4 = 2$ with a remainder of 3.

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

  The value is $\boxed{-i}$.

• c) $i^{42}$ Divide the exponent by 4: $42 \div 4 = 10$ with a remainder of 2.

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

  The value is $\boxed{-1}$.

• d) $i^{105}$ Divide the exponent by 4: $105 \div 4 = 26$ with a remainder of 1.

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

  The value is $\boxed{i}$.