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.

• (1) Please put distinct parts into distinct questions. (2) Please use MathJax to format your math, it makes it a lot easier to read. Apr 16 '17 at 21:35
• @MichaelBurr Sorry, I do not know how to use MathJax, I tried my best to put it in the format. I labeled the parts a-c, but I will rename to make is clearer. Apr 16 '17 at 21:38
• What I mean is that if you have three parts, you should have three questions (one question for part $(a)$, one question for part $(b)$, and one question for part $(c)$). Apr 16 '17 at 21:39
• @MichaelBurr Okay, noted, thank you. It is just that is all one problem that has three parts and I wanted to see if my part a and b are okay to then get help for c. Apr 16 '17 at 21:40
• Also, to learn about MathJax, see the tutorial here. Apr 16 '17 at 21:41

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.

• Fixed now, thanks a lot. The way it was before, I did not like it either, but I did not know there is also \bar. Apr 16 '17 at 22:21
• Personally, I often use \overline because \bar is intended to be used as accent, but in this particular case overline looks plain ugly. Apr 16 '17 at 22:26

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 $N(v^k)=N(v)^k$.

• For $N(v)=N(\overline{v})$ I know I will get a$^2$+b$^2$ for v, but how can I just plug in $\overline{v}$ into the equation if this is for v not $\overline{v}$? Apr 16 '17 at 21:43
• You have a general formula, it might be confusing because you're using $a$ and $b$ in two different roles, I'll edit to make this more clear. Apr 16 '17 at 21:44
• Okay so when plugging in $\overline{w}=c-di$ into the given equation for (a), what exactly is happening? I am not sure how you got (-d)$^2$ because wouldn't this be showing the equations are not equal? Apr 16 '17 at 21:50
• What is the square of a negative real number? Apr 16 '17 at 21:54
• @Pam_22R This shows that the terms are equal since $(-d)^2=-d\cdot (-d)=d^2$. Apr 16 '17 at 21:55