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

In summary: 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) +...
  • #1
chwala
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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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  • #2
If you have already proven |zw|=|z||w| then you can use that fact here to make the proof very short.
 
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  • #3
chwala said:
A different approach would be appreciated...
... and definitely needed.

What about using the properties of the modulus of a complex number?
 
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  • #4
but one will still need to do some computation on the rhs ...i.e if
##z=c+di##, then ##⇒|z|=(c+di)^2##
 
  • #5
chwala said:
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$$|z|^2 = zz^*$$
 
  • #6
PeroK said:
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...
 
  • #7
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.
 
  • #8
Ibix said:
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!
 
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  • #9
Ibix said:
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?
 
  • #10
chwala said:
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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  • #11
PeroK said:
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; $$\frac{|z|}{|z^{*}|}=\frac{|a^2+b^2|}{|a^2-b^2|}=\frac{a^2+b^2}{a^2+b^2}=1=c^2+d^2$$
 
  • #12
If ##z/z^* = c+di##, then ##(z/z^*)^* = z^*/z = c - di##. It follows that
$$
1= \frac{z}{z^*} \frac{z^*}{z} = (c+di)(c-di) = c^2 + d^2.
$$
 
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  • #13
Orodruin said:
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:biggrin:...this was straightforward and directly to the point, ...i need to refresh on the complex number properties...
 
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  • #14
Ibix said:
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...
 
  • #15
chwala said:
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##.
 
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  • #16
[tex]c^2+d^2=1[/tex]
means
[tex]|\frac{z}{z^*}|=1[/tex]
which is obviously true.
 
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  • #17
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.
 
  • #18
PeroK said:
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.
 
  • #19
Ibix said:
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)##.
 
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  • #20
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
$$
z=zz_0 z_0^{*}
$$
$$
z_0 z_0^{*}=1=c^2+d^2
$$
 
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  • #21
Ibix said:
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.
 
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FAQ: Prove that ##c^2+d^2=1## in the problem involving complex numbers

1. What is the purpose of proving that ##c^2+d^2=1## in the problem involving complex numbers?

The purpose of proving this equation is to show that the sum of the squares of two complex numbers, c and d, is equal to 1. This is an important property of complex numbers and can be used in various mathematical applications and calculations.

2. How do you prove that ##c^2+d^2=1## in the problem involving complex numbers?

To prove this equation, we can use the Pythagorean theorem in the complex plane. We can represent c and d as points on the complex plane, and the distance from the origin to these points will be their absolute values, |c| and |d|. By using the Pythagorean theorem, we can show that |c|^2 + |d|^2 = 1, which proves the equation.

3. Can you provide an example of how to use ##c^2+d^2=1## in a mathematical calculation?

Sure, let's say we have two complex numbers, c = 3 + 4i and d = 2 + 5i. We can calculate their absolute values as |c| = √(3^2 + 4^2) = 5 and |d| = √(2^2 + 5^2) = √29. By plugging these values into the equation |c|^2 + |d|^2 = 1, we get 5^2 + (√29)^2 = 1, which is true and proves the equation.

4. Is the equation ##c^2+d^2=1## only valid for complex numbers?

Yes, this equation is only valid for complex numbers. It cannot be applied to real numbers or other types of numbers.

5. What other properties of complex numbers can be proven using ##c^2+d^2=1##?

This equation is just one of the many properties of complex numbers that can be proven. It can also be used to prove the Pythagorean theorem in the complex plane, as well as other trigonometric identities involving complex numbers. It is a fundamental property that is often used in more complex mathematical proofs and calculations involving complex numbers.

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