$(a+bi)$ of $\frac{3+4i}{5+6i}$ and $i^{2006}$ and $\frac{1}{\frac{1}{1+i}-1}$

Question

How can one get the Cartesian coordinate form $(a+bi)$ of the following complex numbers?

$$\frac{3+4i}{5+6i}$$ $$i^{2006}$$ $$\frac{1}{\frac{1}{1+i}-1}$$

Regarding $\frac{3+4i}{5+6i}$ I tried expanding it with $\frac{5+6i}{5+6i}$ and got $\frac{-9+38i}{-9+60i}$, but that doesn't get me anywhere.

Regarding $i^{2006}$ I have $i^{2006} = -1$ (because $2006$ is an even number and $i^2 = -1$). The Cartesian coordinate form would then be $-1 + 0i$. So the real part is $-1$ and the imaginary is $0i$ is that correct?

Regarding $\frac{1}{\frac{1}{1+i}-1}$ I tried expanding the denominator with $\frac{1+i}{1+i}$ and got $\frac{1}{\frac{1}{1+i}-\frac{1+i}{1+i}} = \frac{\frac{1}{1}}{\frac{1+i}{i^2}} = \frac{-1}{1+i}$, but how do I continue from there?

Answer 1

For $\frac{3+4i}{5+6i}$ multiply the numerator and denominator by the conjugate of the denominator, $5 - 6i$, what do you observe ?

For $i^{2006}$. If $i^2 = -1$, what can you say about $i^{2006} = (i^2)^{1003} = (-1)^{1003} $ ?

For $\frac{1}{\frac{1}{1+i}-1}$, multiply numerator and denominator by $1+i$, you get $\frac{1+i}{-i} = -\frac{1}{i} - 1 = -1 +i$

Answer 2

$$ \frac{z_1}{z_2}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} $$

$$\frac{z_1}{z_2}=(\frac{r_1}{r_2})e^{i(\theta_1-\theta_2)}$$