How do I prove this?

[Immortal]

Hai, I was trying to solve previous years exam question papers and I came across this question.

Q) Given that for any complex numbers z, |z|² =z z^{- (the second z after = sign is z conjugate, I cant type it properly)}

i) prove that |z₁+z₂|²=|z₁|²+2Re(z1 z₂) + |z₂|² (the z₂ after the 2Re is z₂​ conjugate , I cant type it )
ii) Prove that |z₁+z₂|² +|z₁-z₂|²=2[|z₁|²+|z₂|²]

Where z₁ and z₂ are any two complex numbers.

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How do I prove this?
In this lesson, I've never came across a problem like this. I know (z1+z2)² = z1²+2z1z2+z2² I applied that formula and the equation became
|z1²+2z1 z2+ z2²| How do I progress further?

Then I thought like this ,
assume z1= a+ib and z2= c+id
so 2 z1 z2 =2(a+ib)(c+id)
= 2 (ac-bd) +i(ad+bc)
And...then I have no idea what to do next.

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In the second subquestion LHS becomes
|z1²+2z1z2+z2²| + |z1²-2z1z2+z2²| after applying the formula of (a+b)²

+2z1z2 and -2z1z2 gets cancelled and the remaining is |z1²+z2²|+|z1²+z2²|
I.e 2{|z1²+z2²|}
I dont know what to do next.
Is |z1²| + |z2|² same as |z1²+z2²​|?

 

[pka]

Hai, I was trying to solve previous years exam question papers and I came across this question.
Q) Given that for any complex numbers z, |z|² =z z^{- (the second z after = sign is z conjugate, I cant type it properly)}
i) prove that |z₁+z₂|²=|z₁|²+2Re(z1 z₂) + |z₂|² (the z₂ after the 2Re is z₂​ conjugate , I cant type it )
ii) Prove that |z₁+z₂|² +|z₁-z₂|²=2[|z₁|²+|z₂|²]

First let's not use subscripts: say that each of \(\displaystyle z~\&~w\) is a complex number.
In addition to \(\displaystyle |z|^2=z\cdot\overline{z}\) you need to know:

\(\displaystyle \overline{z+w}=\overline{z}+\overline{w},~~z+ \overline{z}=2\mathfrak{R}(z)~\&~z- \overline{z}=2\mathfrak{I}(z)\)

Thus \(\displaystyle |z+w|^2=(z+w)\cdot\overline{z+w}=(z+w)\cdot ( \overline{z}+\overline{w})=~?\)

Now YOU finish.

 

[Immortal]

Thus \(\displaystyle |z+w|^2=(z+w)\cdot\overline{z+w}=(z+w)\cdot ( \overline{z}+\overline{w})=~?\)

Now YOU finish.

Those formulas are not there in our syllabus but erhm..
So the LHS becomes

\(\displaystyle |z+w|^2=(z+w)\cdot\overline{z+w} \)
\(\displaystyle =(z+w)(\overline{z}+\overline{w}) \)
\(\displaystyle =z\overline{z}+z\overline{w}+w\overline{z} + w\overline{w}\)
\(\displaystyle =|z|^2+2Re(z\overline{w})+|w|^2 \)

Now the 2nd question,
By using those formulas, we get
\(\displaystyle |z+w|^2=z\overline{z}+z\overline{w}+w\overline{z} + w\overline{w}\)
and \(\displaystyle |z-w|^2=z\overline{z}-z\overline{w}-w\overline{z} + w\overline{w} \)
So when we add, the \(\displaystyle +z\overline{w}+w\overline{z} and -z\overline{w}-w\overline{z} \)gets cancelled and whats remaining will be
\(\displaystyle 2|z|^2+2|w|^2 \)
The 2 is common so by taking it outside, we get
\(\displaystyle 2[|z|^2+|w|^2]\)
Hence proved.
I guess I didn't do anything wrong.
Thanks, But I've no idea why they asked questions involving formulas which isn't included in our syllabus in exam.
I better do some research on all the formulas related to complex numbers.
One more question which is out of topic,
while I was waiting for someone to reply to my question, I went through other's questions and I knew the answer of some of them. Is it allowed for new members to provide answers?

 

[Deleted member 4993]

One more question which is out of topic,
while I was waiting for someone to reply to my question, I went through other's questions and I knew the answer of some of them. Is it allowed for new members to provide answers?

It is perfectly alright for anyone to answer any question in this forum.

However, if you noticed, we refrain from providing complete answers. You may do an example problem completly to show the flow of logic.

Leave something for the poster to discover.