# Problem 1: Section 1.1 #10

To Write $$\left( \frac{2+i}{6i-(1-2i)} \right) ^2$$, we must first expand it by multiplying it by itself, as so: $\frac{2+i}{6i-(1-2i)}*\frac{2+i}{6i-(1-2i)} = \frac{4+4i-1}{-64-16i+1} = \frac{3+4i}{-(63+16i)}$
and then multiplying by its complex conjugate to obtain a separable real and imaginary part: $\frac{3+4i}{-(63+16i)}*\frac{-(63-16i)}{-(63-16i)} = \frac{189-48i+252i+64}{-(63^2+16^2)}$ which simplifies to $- \left( \frac{253}{4225}+\frac{204}{4225}i \right)$

# Problem 2: Section 1.1 #14

We must show $$Re(iz) = -im(z), \forall z \in \mathbb {C}$$
To show this, we will examine both sides of this equation in the context that every complex number z can be expressed as $$z=a+ib$$. Looking first at the left hand side: $Re(iz) = Re(i(a+ib)) = Re(ia-b) = -b$
Looking at the right hand side: $-im(z) = -im (a+ib) =-b$ So one can see that $Re(iz) = -b = -im(z)$ implying that $Re(iz) = -im(z)$

# Problem 3: Section 1.1 #15

## Part a)

We wish to show: $$i^{4k}=1, k \in \mathbb {z}$$
This can be easily seen if we use the definition of i: $$i=\sqrt{-i}$$ which implies that $$i^{4k} = (-1)^{2k} = 1^k = 1$$

## Part b)

We wish to show: $$i^{4k+1} = i, k \in \mathbb {z}$$
We also want to use the definion of i here, but first we want to manipulate the exponent as $$i^{4k+1} = i^{4k}i^1$$ and then using the definition: $i^{4k}i^1 = \sqrt{-1}^{4k}i = (-1)^{2k}i = 1^{k}i = i$ $\therefore i^{4k+1} = i$

## part c)

We wish to show: $$i^{4k+2} = -1, k \in \mathbb {z}$$
We will use a similar method as in b: $i^{4k+2} = i^{4k}i^2 = \sqrt{-1}^{4k}\sqrt{-1}^{2} = (-1)^{2k} * (-1) = 1^k * -1 = -1$ $\therefore i^{4k+2} = -1$

## part d)

We wish to show: $$i^{4k+3} = -i, k \in \mathbb {z}$$
We will use a similar method as in b and c: $i^{4k+3} = i^{4k}i^3 = \sqrt{-1}^{4k}i^2i = (-1)^{2k}\sqrt{-1}^2i = 1^k*-1i = -i$ $\therefore i^{4k+3} = -i$