# Prove the given complex number problem

• chwala

#### chwala

Gold Member
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...

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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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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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• chwala