Question

If $z_1$ and $z_2$ are complex numbers, prove that $|z_1+z_2|^2$ = $|z_1|^2+|z_2|^2$ if and only if \[{z_1}\mathop {{z_2}}\limits^\_ \] z is pure imaginary.

Answer 1

Hint:Proceed by opening the square of the term in the LHS. Use identities and open it then observe the results to prove the statement.
Identities:
 $ |{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} + 2\operatorname{Re} |{z_1}{z_2}| $
 $ 2\operatorname{Re} |{z_1}{z_2}| = {z_1}\mathop {{z_2}}\limits^\_ + \mathop {{z_1}}\limits^\_ {z_2} $

Complete step-by-step answer:
To prove:
$|z_1+z_2|^2$ = $|z_1|^2+|z_2|^2$ if and only if \[{z_1}\mathop {{z_2}}\limits^\_ \]
Opening the square on the LHS, using the identity::
 $ |{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} + 2\operatorname{Re} |{z_1}{z_2}| $
We get,
 $ |{z_1}{|^2} + |{z_2}{|^2} + 2\operatorname{Re} |{z_1}{z_2}| = |{z_1}{|^2} + |{z_2}{|^2} $
Cancelling the same terms, gives:
 $ 2\operatorname{Re} |{z_1}{z_2}| = 0 $
 $ 2\operatorname{Re} |{z_1}{z_2}| = {z_1}\mathop {{z_2}}\limits^\_ + \mathop {{z_1}}\limits^\_ {z_2} = 0 $
$
  {z_1}\mathop {{z_2}}\limits^\_ = - \mathop {{z_1}}\limits^\_ {z_2} \\
  {z_1}\mathop {{z_2}}\limits^\_ + \mathop {{z_1}}\limits^\_ {z_2} = 0 \\
$
This implies that $ \operatorname{Re} ({z_1}\mathop {{z_2})}\limits^\_ = 0 $
If the real part of this complex is zero then it means that the complex number is purely imaginary.

Note:Other conclusions which can derived from this result are:
 $ \dfrac{{{z_1}}}{{{{\mathop z\limits^\_ }_2}}} $ is also purely imaginary and \[{z_1}\mathop {{z_2}}\limits^\_ + \mathop {{z_1}}\limits^\_ {z_2} = 0\].
These results are a direct implication of the above proof, so they can be remembered.