If ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers such that ${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}$, then:
(a) \[{{z}_{1}}\overline{{{z}_{2}}}\] is purely imaginary.
(b) \[\dfrac{{{z}_{1}}}{{{z}_{2}}}\] is purely imaginary.
(c) \[{{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=0\]
(d) $O,{{z}_{1}},{{z}_{2}}$ are vertices of a right angle triangle.
Answer
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Hint: Assume ${{z}_{1}}={{x}_{1}}+i{{y}_{1}}$ and ${{z}_{2}}={{x}_{2}}+i{{y}_{2}}$ as the two complex numbers where ${{x}_{1}},{{y}_{1}}$ are the real, imaginary part if ${{z}_{1}}$ respectively and ${{x}_{2}},{{y}_{2}}$ are the real and imaginary part of ${{z}_{2}}$ respectively. Now, write their conjugates as $\overline{{{z}_{1}}}={{x}_{1}}-i{{y}_{1}}$ and $\overline{{{z}_{2}}}={{x}_{2}}-i{{y}_{2}}$. Using the relation given between the modulus of these complex numbers form a relation between ${{x}_{1}},{{y}_{1}},{{x}_{2}},{{y}_{2}}$. Use the formula ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}$. Now, check each option one by one. For option (a) take the product \[{{z}_{1}}\overline{{{z}_{2}}}\] and see if their real part is 0. For option (b) use the result obtained in (a). For option (c) solve the L.H.S and see if it is equal to 0 in the R.H.S. for option (d) use the result obtained in option (b) to prove that $\arg \left( {{z}_{1}} \right)-\arg \left( {{z}_{2}} \right)=\dfrac{\pi }{2}$ by using the formulas $arg\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\arg \left( {{z}_{1}} \right)-\arg \left( {{z}_{2}} \right)$ and $\arg \left( z \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( z \right)}{\operatorname{Re}\left( z \right)} \right)$.
Complete step by step solution:
Here we have been provided with two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$ with the relation ${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}$. We are to find the correct option(s).
Now, let use assume these complex numbers as ${{z}_{1}}={{x}_{1}}+i{{y}_{1}}$ and ${{z}_{2}}={{x}_{2}}+i{{y}_{2}}$ where ${{x}_{1}},{{y}_{1}}$ are the real, imaginary part if ${{z}_{1}}$ respectively and ${{x}_{2}},{{y}_{2}}$ are the real and imaginary part of ${{z}_{2}}$ respectively. We know that modulus of a complex number is given as ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}$, so considering the given relation we get,
\[\begin{align}
& \Rightarrow {{\left| \left( {{x}_{1}}+i{{y}_{1}} \right)+\left( {{x}_{2}}+i{{y}_{2}} \right) \right|}^{2}}={{\left| \left( {{x}_{1}}+i{{y}_{1}} \right) \right|}^{2}}+{{\left| \left( {{x}_{2}}+i{{y}_{2}} \right) \right|}^{2}} \\
& \Rightarrow {{\left| \left( {{x}_{1}}+{{x}_{2}} \right)+i\left( {{y}_{1}}+{{y}_{2}} \right) \right|}^{2}}={{\left| \left( {{x}_{1}}+i{{y}_{1}} \right) \right|}^{2}}+{{\left| \left( {{x}_{2}}+i{{y}_{2}} \right) \right|}^{2}} \\
& \Rightarrow {{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}=\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)+\left( {{x}_{2}}^{2}+{{y}_{2}}^{2} \right) \\
\end{align}\]
Expanding the L.H.S by using the algebraic identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] we get,
\[\begin{align}
& \Rightarrow {{x}_{1}}^{2}+{{x}_{2}}^{2}+2{{x}_{1}}{{x}_{2}}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+2{{y}_{1}}{{y}_{2}}={{x}_{1}}^{2}+{{y}_{1}}^{2}+{{x}_{2}}^{2}+{{y}_{2}}^{2} \\
& \Rightarrow 2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)=0 \\
& \Rightarrow \left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)=0...............\left( i \right) \\
\end{align}\]
Now, let us check each option one by one.
(a) Here we have to consider the product \[{{z}_{1}}\overline{{{z}_{2}}}\], where \[\overline{z}\] denotes the conjugate of $z$ obtained by changing the sign between the real and imaginary part of $z$, and check if it is purely imaginary. This product will be purely imaginary only if the real part will be 0 and there will be only imaginary part.
\[\begin{align}
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}=\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right) \\
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}={{x}_{1}}{{x}_{2}}-i{{x}_{1}}{{y}_{2}}+i{{y}_{1}}{{x}_{2}}-{{i}^{2}}{{y}_{1}}{{y}_{2}} \\
\end{align}\]
We know that \[{{i}^{2}}=-1\] so we get,
\[\begin{align}
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}={{x}_{1}}{{x}_{2}}-i{{x}_{1}}{{y}_{2}}+i{{y}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \\
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}=\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right) \\
\end{align}\]
Using relation (i) we get,
\[\therefore {{z}_{1}}\overline{{{z}_{2}}}=-i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right)\]
Clearly we can see that here we have only imaginary part so the product \[{{z}_{1}}\overline{{{z}_{2}}}\] is purely imaginary.
(b) Here we have to consider the relation \[\dfrac{{{z}_{1}}}{{{z}_{2}}}\] and check if this is purely imaginary or not.
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left( {{x}_{1}}+i{{y}_{1}} \right)}{\left( {{x}_{2}}+i{{y}_{2}} \right)}\]
Rationalizing the denominator by multiplying and dividing with \[\left( {{x}_{2}}-i{{y}_{2}} \right)\] we get,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left( {{x}_{1}}+i{{y}_{1}} \right)}{\left( {{x}_{2}}+i{{y}_{2}} \right)}\times \dfrac{\left( {{x}_{2}}-i{{y}_{2}} \right)}{\left( {{x}_{2}}-i{{y}_{2}} \right)}\]
We can write \[\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)={{z}_{1}}\overline{{{z}_{2}}}\] and \[\left( {{x}_{2}}+i{{y}_{2}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)={{\left| {{z}_{2}} \right|}^{2}}\] so we get,
\[\therefore \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{z}_{1}}\overline{{{z}_{2}}}}{{{x}_{2}}^{2}+{{y}_{2}}^{2}}\]
Clearly we can see that here the denominator is real and by using the conclusion of option (a) numerator is imaginary so \[\dfrac{{{z}_{1}}}{{{z}_{2}}}\] will be purely imaginary.
(c) Here we have to check the validation of the relation \[{{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=0\]. So let us simplify the L.H.S.
\[\Rightarrow {{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)+\left( {{x}_{1}}-i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right)\]
On simplifying and using relation (i) and the relation obtained in option (a) we get,
\[\begin{align}
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=-i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right)+i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right) \\
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=0 \\
& \therefore L.H.S=R.H.S \\
\end{align}\]
(d) Here we have to if $O,{{z}_{1}},{{z}_{2}}$ are vertices of a right angle triangle. This condition will be fulfilled only when the angle between the line joining the O, ${{z}_{1}}$ and the line joining O, ${{z}_{2}}$ in the argand plane will be $\dfrac{\pi }{2}$. In other words, the difference between the arguments of ${{z}_{1}}$ and ${{z}_{2}}$ should be $\dfrac{\pi }{2}$ where $\arg \left( z \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( z \right)}{\operatorname{Re}\left( z \right)} \right)$. Here Im (z) and Re (z) denotes the imaginary part and real part of z respectively.
Using the result of option (b) we have we can say that $\operatorname{Re}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=0$ so we have,
\[\begin{align}
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}{\operatorname{Re}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)} \right) \\
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}{0} \right) \\
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\infty \\
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\dfrac{\pi }{2} \\
\end{align}\]
Using the formula $arg\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\arg \left( {{z}_{1}} \right)-\arg \left( {{z}_{2}} \right)$ we get,
$\therefore \arg \left( {{z}_{1}} \right)-\arg \left( {{z}_{2}} \right)=\dfrac{\pi }{2}$
Therefore angle between $O{{z}_{1}}$ and $O{{z}_{2}}$ is 90 degrees so $O,{{z}_{1}},{{z}_{2}}$ are the vertices of the right triangle.
Hence all the options are correct.
Note: Remember the basic terms of complex numbers like conjugate, argument, modulus etc. We do not represent complex numbers on a real plane but there is a complex plane for their representation. Note that if the imaginary part of a complex number is 0 then it is called purely real. Remember the formulas of argument of complex numbers as they are helpful in solving angle based problems.
Complete step by step solution:
Here we have been provided with two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$ with the relation ${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}$. We are to find the correct option(s).
Now, let use assume these complex numbers as ${{z}_{1}}={{x}_{1}}+i{{y}_{1}}$ and ${{z}_{2}}={{x}_{2}}+i{{y}_{2}}$ where ${{x}_{1}},{{y}_{1}}$ are the real, imaginary part if ${{z}_{1}}$ respectively and ${{x}_{2}},{{y}_{2}}$ are the real and imaginary part of ${{z}_{2}}$ respectively. We know that modulus of a complex number is given as ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}$, so considering the given relation we get,
\[\begin{align}
& \Rightarrow {{\left| \left( {{x}_{1}}+i{{y}_{1}} \right)+\left( {{x}_{2}}+i{{y}_{2}} \right) \right|}^{2}}={{\left| \left( {{x}_{1}}+i{{y}_{1}} \right) \right|}^{2}}+{{\left| \left( {{x}_{2}}+i{{y}_{2}} \right) \right|}^{2}} \\
& \Rightarrow {{\left| \left( {{x}_{1}}+{{x}_{2}} \right)+i\left( {{y}_{1}}+{{y}_{2}} \right) \right|}^{2}}={{\left| \left( {{x}_{1}}+i{{y}_{1}} \right) \right|}^{2}}+{{\left| \left( {{x}_{2}}+i{{y}_{2}} \right) \right|}^{2}} \\
& \Rightarrow {{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}=\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)+\left( {{x}_{2}}^{2}+{{y}_{2}}^{2} \right) \\
\end{align}\]
Expanding the L.H.S by using the algebraic identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] we get,
\[\begin{align}
& \Rightarrow {{x}_{1}}^{2}+{{x}_{2}}^{2}+2{{x}_{1}}{{x}_{2}}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+2{{y}_{1}}{{y}_{2}}={{x}_{1}}^{2}+{{y}_{1}}^{2}+{{x}_{2}}^{2}+{{y}_{2}}^{2} \\
& \Rightarrow 2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)=0 \\
& \Rightarrow \left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)=0...............\left( i \right) \\
\end{align}\]
Now, let us check each option one by one.
(a) Here we have to consider the product \[{{z}_{1}}\overline{{{z}_{2}}}\], where \[\overline{z}\] denotes the conjugate of $z$ obtained by changing the sign between the real and imaginary part of $z$, and check if it is purely imaginary. This product will be purely imaginary only if the real part will be 0 and there will be only imaginary part.
\[\begin{align}
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}=\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right) \\
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}={{x}_{1}}{{x}_{2}}-i{{x}_{1}}{{y}_{2}}+i{{y}_{1}}{{x}_{2}}-{{i}^{2}}{{y}_{1}}{{y}_{2}} \\
\end{align}\]
We know that \[{{i}^{2}}=-1\] so we get,
\[\begin{align}
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}={{x}_{1}}{{x}_{2}}-i{{x}_{1}}{{y}_{2}}+i{{y}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \\
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}=\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right) \\
\end{align}\]
Using relation (i) we get,
\[\therefore {{z}_{1}}\overline{{{z}_{2}}}=-i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right)\]
Clearly we can see that here we have only imaginary part so the product \[{{z}_{1}}\overline{{{z}_{2}}}\] is purely imaginary.
(b) Here we have to consider the relation \[\dfrac{{{z}_{1}}}{{{z}_{2}}}\] and check if this is purely imaginary or not.
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left( {{x}_{1}}+i{{y}_{1}} \right)}{\left( {{x}_{2}}+i{{y}_{2}} \right)}\]
Rationalizing the denominator by multiplying and dividing with \[\left( {{x}_{2}}-i{{y}_{2}} \right)\] we get,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left( {{x}_{1}}+i{{y}_{1}} \right)}{\left( {{x}_{2}}+i{{y}_{2}} \right)}\times \dfrac{\left( {{x}_{2}}-i{{y}_{2}} \right)}{\left( {{x}_{2}}-i{{y}_{2}} \right)}\]
We can write \[\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)={{z}_{1}}\overline{{{z}_{2}}}\] and \[\left( {{x}_{2}}+i{{y}_{2}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)={{\left| {{z}_{2}} \right|}^{2}}\] so we get,
\[\therefore \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{z}_{1}}\overline{{{z}_{2}}}}{{{x}_{2}}^{2}+{{y}_{2}}^{2}}\]
Clearly we can see that here the denominator is real and by using the conclusion of option (a) numerator is imaginary so \[\dfrac{{{z}_{1}}}{{{z}_{2}}}\] will be purely imaginary.
(c) Here we have to check the validation of the relation \[{{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=0\]. So let us simplify the L.H.S.
\[\Rightarrow {{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}-i{{y}_{2}} \right)+\left( {{x}_{1}}-i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right)\]
On simplifying and using relation (i) and the relation obtained in option (a) we get,
\[\begin{align}
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=-i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right)+i\left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right) \\
& \Rightarrow {{z}_{1}}\overline{{{z}_{2}}}+\overline{{{z}_{1}}}{{z}_{2}}=0 \\
& \therefore L.H.S=R.H.S \\
\end{align}\]
(d) Here we have to if $O,{{z}_{1}},{{z}_{2}}$ are vertices of a right angle triangle. This condition will be fulfilled only when the angle between the line joining the O, ${{z}_{1}}$ and the line joining O, ${{z}_{2}}$ in the argand plane will be $\dfrac{\pi }{2}$. In other words, the difference between the arguments of ${{z}_{1}}$ and ${{z}_{2}}$ should be $\dfrac{\pi }{2}$ where $\arg \left( z \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( z \right)}{\operatorname{Re}\left( z \right)} \right)$. Here Im (z) and Re (z) denotes the imaginary part and real part of z respectively.
Using the result of option (b) we have we can say that $\operatorname{Re}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=0$ so we have,
\[\begin{align}
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}{\operatorname{Re}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)} \right) \\
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}{0} \right) \\
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\infty \\
& \Rightarrow \arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\dfrac{\pi }{2} \\
\end{align}\]
Using the formula $arg\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\arg \left( {{z}_{1}} \right)-\arg \left( {{z}_{2}} \right)$ we get,
$\therefore \arg \left( {{z}_{1}} \right)-\arg \left( {{z}_{2}} \right)=\dfrac{\pi }{2}$
Therefore angle between $O{{z}_{1}}$ and $O{{z}_{2}}$ is 90 degrees so $O,{{z}_{1}},{{z}_{2}}$ are the vertices of the right triangle.
Hence all the options are correct.
Note: Remember the basic terms of complex numbers like conjugate, argument, modulus etc. We do not represent complex numbers on a real plane but there is a complex plane for their representation. Note that if the imaginary part of a complex number is 0 then it is called purely real. Remember the formulas of argument of complex numbers as they are helpful in solving angle based problems.
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