Prove that ##c^2+d^2=1## in the problem involving complex numbers

[chwala]

Homework Statement

Let $z=a+bi$, where $a$ and $b$ are real numbers. If Let $z=a+bi$, where $a$ and $b$ are real numbers. If $$\frac {z}{z^*}=c+di$$, where $c$ and $d$ are real, prove that $c^2+d^2=1$

Relevant Equations

Complex numbers

Easy questions, just a lot of computation...

$$\frac {z}{z^*}=\frac {a+bi}{a-bi} ×\frac {a+bi}{a+bi}$$
$$c+di=\frac {a^2-b^2}{a^2+b^2}+\frac {2abi}{a^2+b^2}$$
$$⇒c^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}$$
$$⇒d^2= \frac {4a^2b^2}{(a^2+b^2)^2}$$
Therefore, $$c^2+d^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}+\frac {4a^2b^2}{(a^2+b^2)^2}=1$$

A different approach would be appreciated...

 

[Office_Shredder]

If you have already proven |zw|=|z||w| then you can use that fact here to make the proof very short.

 

[PeroK]

A different approach would be appreciated...

... and definitely needed.

What about using the properties of the modulus of a complex number?

 

[chwala]

but one will still need to do some computation on the rhs ...i.e if
$z=c+di$, then $⇒|z|=(c+di)^2$

 

[PeroK]

but one will still need to do some computation on the rhs ...i.e if
$z=c+di$, then $⇒|z|=(c+di)^2$

That's not right. If $z = a +bi$, then$$|z|^2 = a^2 + b^2$$And, in fact, we also have$$|z|^2 = zz^*$$

 

[chwala]

That's not right. If $z = a +bi$, then$$|z|^2 = a^2 + b^2$$And, in fact, we also have$$|z|^2 = zz^*$$

Noted, i made a mistake there...

 

[Ibix]

Are you aware that you can write a complex number $z=a+ib$ as $|z|e^{i\theta}$, where $\tan\theta=b/a$? If so, it's really easy.

 

[PeroK]

Are you aware that you can write a complex number $z=a+ib$ as $|z|e^{i\theta}$, where $\tan\theta=b/a$? If so, it's really easy.

Not as easy as it is using the modulus!

 

[chwala]

Are you aware that you can write a complex number $z=a+ib$ as $|z|e^{i\theta}$, where $\tan\theta=b/a$? If so, it's really easy.

I need to refresh on this...yes i am aware...are you talking of the Euler form of equation...something like
$$z=x+iy= r{cos θ+i sin θ}$$...applying $z^n$ where necessary?

 

[Ibix]

I need to refresh on this...yes i am aware...you're talking of the euler form something like
$$z=x+iy= r{cos θ+i sin θ}$$...applying $z^n$ where necessary?

Yeah, but if you express it the way I did with the complex exponential and note that $z^*=|z|e^{-i\theta}$, it should be a one-liner. This approach implies the result about moduli that I think @PeroK is advocating using directly.

Edit: by the way, you've got a LaTeX bug - you tried to use {} instead of (). Braces aren't rendered, though, so your $r$ appears to be multiplying only the $\cos$ instead of the $\cos$ and the $i\sin$.

 

[chwala]

That's not right. If $z = a +bi$, then$$|z|^2 = a^2 + b^2$$And, in fact, we also have$$|z|^2 = zz^*$$

Ok going with this thinking, we shall get; $$\frac{|z|}{|z^{*}|}=\frac{|a^2+b^2|}{|a^2-b^2|}=\frac{a^2+b^2}{a^2+b^2}=1=c^2+d^2$$

 

[Orodruin]

If $z/z^* = c+di$, then $(z/z^*)^* = z^*/z = c - di$. It follows that
$$
1= \frac{z}{z^*} \frac{z^*}{z} = (c+di)(c-di) = c^2 + d^2.
$$

 

[chwala]

If $z/z^* = c+di$, then $(z/z^*)^* = z^*/z = c - di$. It follows that
$$
1= \frac{z}{z^*} \frac{z^*}{z} = (c+di)(c-di) = c^2 + d^2.
$$

Nice one mate...this was straightforward and directly to the point, ...i need to refresh on the complex number properties...

 

[chwala]

Are you aware that you can write a complex number $z=a+ib$ as $|z|e^{i\theta}$, where $\tan\theta=b/a$? If so, it's really easy.

How would your solution look like, you're transforming to polar form of equation? How will you treat the argument? Cheers...

 

[Ibix]

How would your solution look like, you're transforming to polar form of equation? How will you treat the argument? Cheers...

$$\frac{z}{z^*}=\frac{|z|e^{i\theta}}{|z|e^{-i\theta}}=e^{2i\theta}$$That last expression is a complex number with unit modulus written in complex exponential form and must be equal to $c+id$. Hence $|c+id|=1$.

 

[anuttarasammyak]

$$c^2+d^2=1$$
means
$$|\frac{z}{z^*}|=1$$
which is obviously true.

 

[PeroK]

My solution would be simply:
$$c^2 + d^2 = \big |\frac z {z^*}\big |^2 = \frac{|z|^2}{|z^*|^2} = 1$$I can't see that the polar form is needed.

 

[Ibix]

I can't see that the polar form is needed.

Depends whether you remember that $|ab|=|a||b|$ off the top of your head or not. It's one of those things I never quite recall for some reason, so going via $ab=|ab|\exp(i(\theta_a+\theta_b))$ is easier for me.

 

[PeroK]

Depends whether you remember that $|ab|=|a||b|$ off the top of your head or not. It's one of those things I never quite recall for some reason, so going via $ab=|ab|\exp(i(\theta_a+\theta_b))$ is easier for me.

Well, if you can remember that $z = |z|e^{i\theta}$, then $$zw = |z|e^{i\theta}|w|e^{i\phi} = |z||w|e^{i(\theta + \phi)}$$From which is follows that $|zw| = |z||w|$ and $\arg(zw) = \arg(z) + \arg(w)$.

 

[Fred Wright]

Another way. Let $z_0=c+id$. We have,
$$
z=z^{*}z_0
$$
Taking the complex conjugate of both sides of the equation,
$$
z^{*}=zz_0^{*}
$$
substituting $z^{*}$ in the first equation
$$
z=zz_0 z_0^{*}
$$
$$
z_0 z_0^{*}=1=c^2+d^2
$$

 

[Orodruin]

Depends whether you remember that $|ab|=|a||b|$ off the top of your head or not.

You do not need to remember that. Nor that $c^2 + d^2$ is the norm squared of $c+di$. All you need to know is how complex conjugation works on multiplications. See #12. Both #17 and #20 are rather minor variations of #12.