Complex Numbers and the Norm I am given that for a complex number $w=a+bi$, define $\overline{w}=a-bi$ and $N(w)=w \overline{w}$. I have provided my answers for parts a and b, but I am not sure they are correct. I need help with figuring out part c.
(a) Compute $N(w)=w \overline{w}$ explicitly. 
Here is what I have gotten:$N(w)=w \overline{w}= (a+bi)(a-bi)= a^2+b^2$.
(b) Show that $N(rs)=N(r)N(s)$ for any two complex numbers $r,s$.
Here is my work:
Let $r=a+bi$, $s=c+di
$
Then $$rs=(a+bi)(c+di)=ac+adi+cbi-bd=(ac-bd)+(ad+cb)i.
$$
Now, from a, have that $N(r)= a^2+b^2$ and $N(s)= c^2+d^2$. So, $$N(r)N(s)=(a^2+b^2)(c^2+d^2).
$$
Also, $N(rs)=N((ac-bd)+(ad+cb)i) $ and from part a then, $N(rs)=(ac-bd)^2+(ad+cb)^2$
When expanded, 
$$a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2$$ 
\begin{align*}
&=a^2c^2+a^2d^2+b^2d^2+b^2c^2\\
&=a^2(c^2+d^2)+b^2(c^2+d^2)\\
&=(a^2+b^2)(c^2+d^2).
\end{align*}
Hence, $N(rs)=N(r)N(s)$.
(c) Prove that $N(\overline{v})=N(v) $ and $N(v^n)=N(v)^n$ for any complex number.
This is the part I am confused as to how to prove. I let $v=a+bi$ and $\overline{v}=a-bi$. Then I am not sure how to use what I have shown before for this part. 
 A: For the first part of your question, notice that $$N (v)=\bar{v} v=\bar{v} \bar{\bar{v}}=N (\bar{v}) $$
because $\bar{\bar{v}}=v$.
For the second part, consider doing induction over $n $:
Induction basis: $n=1$ $$N (v^1)=N (v)=N (v)^1$$
Induction step: $$N (v^{n+1})=N (v^n\cdot v)=N (v^n) \cdot N (v)=N (v)^n \cdot N (v)=N (v)^{n+1} $$
For the third from last step we used your result from b, for the second from last we used the induction hypothesis.
A: Since your question is about part $(c)$...


*

*To prove $N(v)=N(\overline{v})$, use your formula for $N$ in part $(a)$ on each of $v$ and $\overline{v}$.  As stated, the question is somewhat confusing because $a$ and $b$ are used in two ways in the problem.  In part $(a)$, $N(v)=a^2+b^2$ when $v=a+bi$.  For part $(c)$, we could use $w=c+di$ and $\overline{w}=c-di$, then $N(w)=c^2+d^2$ (by substituting the variables in $w$ into the form for part $(a)$) and $N(\overline{w})=c^2+(-d)^2$ by the same substitution.

*To prove $N(v^n)=N(v)^n$, use your equality in part $(b)$ as well as induction on $n$.  The base case is $N(v^1)=N(v)^1$, and then use part $(b)$ to prove $N(v^{k+1})=N(v)^{k+1}$ using $N(v^k)=N(v)^k$.
