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

• chwala
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) +...f

#### chwala

Gold Member
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...

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

chwala
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##

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$$|z|^2 = zz^*$$

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...

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.

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!

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.
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?

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##.

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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; $$\frac{|z|}{|z^{*}|}=\frac{|a^2+b^2|}{|a^2-b^2|}=\frac{a^2+b^2}{a^2+b^2}=1=c^2+d^2$$

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

PhDeezNutz and 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...

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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...

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##.

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

PeroK and chwala
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.

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.

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)##.

chwala and Ibix
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
$$z=zz_0 z_0^{*}$$
$$z_0 z_0^{*}=1=c^2+d^2$$

chwala
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.

Ibix