Prove the given complex number problem

[chwala]

Homework Statement

Prove that, for any complex number $z$, $zz^{*}= \bigl(\Re (z))^2+\bigl(\Im (z))^2$

Relevant Equations

Complex numbers

This is pretty straightforward,
Let $z=a+bi$
$\bigl(\Re (z))=a, \bigl(\Im (z))=b$
$zz^*=(a+bi)(a-bi)=a^2+b^2 =\bigl(\Re (z))^2+\bigl(\Im (z))^2$
Any other approach? this are pretty simple questions ...all the same its good to explore different perspective on the same...

 

[BvU]

https://www.math-linux.com/latex-26/faq/latex-faq/article/latex-real-part-symbol

$$zz^{*}= \bigl(\Re(z)\bigr )^2+ \bigl(\Im (z)\bigr )^2\qquad {\sf {or}}
\qquad \bigl (\operatorname{Re}(z)\bigr )^2+ \bigl(\operatorname{Im} (z)\bigr )^2$$but I have no trouble admitting both are ugly. Fortunately this doesn't occur all that often. Maybe \Bigr etc is a little better: $$zz^{*}= \Bigl(\Re(z)\Bigr)^2+ \Bigl(\Im (z)\Bigr )^2\qquad {\sf {or}}
\qquad \Bigl (\operatorname{Re}(z)\Bigr )^2+ \Bigl(\operatorname{Im} (z)\Bigr )^2$$

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[chwala]

Thanks Bvu, i have greatly learned latex from you...will take note...

 

[BvU]

Any other approach?

$${\sf z} = |{\sf z}| e^ {i\arg {\sf z} }, \quad {\sf z^*} = |{\sf z}| e^ {-i\arg {\sf z} } \Rightarrow {\sf zz^*} = |{\sf z}|^2 $$but: does that pass muster ?
I guess not ...

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