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,


\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
    $\begingroup$ (1) Please put distinct parts into distinct questions. (2) Please use MathJax to format your math, it makes it a lot easier to read. $\endgroup$ Apr 16 '17 at 21:35
  • $\begingroup$ @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. $\endgroup$
    – Pam_22R
    Apr 16 '17 at 21:38
  • $\begingroup$ 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 $(c)$). $\endgroup$ Apr 16 '17 at 21:39
  • $\begingroup$ @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. $\endgroup$
    – Pam_22R
    Apr 16 '17 at 21:40
  • $\begingroup$ Also, to learn about MathJax, see the tutorial here. $\endgroup$ 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.

  • $\begingroup$ 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. $\endgroup$
    – mxian
    Apr 16 '17 at 22:21
  • $\begingroup$ Personally, I often use \overline because \bar is intended to be used as accent, but in this particular case overline looks plain ugly. $\endgroup$
    – Ennar
    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 $N(v^k)=N(v)^k$.

  • $\begingroup$ 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 $\overline{v}$? $\endgroup$
    – Pam_22R
    Apr 16 '17 at 21:43
  • $\begingroup$ 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. $\endgroup$ Apr 16 '17 at 21:44
  • $\begingroup$ 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? $\endgroup$
    – Pam_22R
    Apr 16 '17 at 21:50
  • $\begingroup$ What is the square of a negative real number? $\endgroup$ Apr 16 '17 at 21:54
  • $\begingroup$ @Pam_22R This shows that the terms are equal since $(-d)^2=-d\cdot (-d)=d^2$. $\endgroup$
    – mxian
    Apr 16 '17 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.