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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 ...

If z1z_1 and z2z_2 are complex numbers, prove that ∣z1+z2∣2|z_1+z_2|^2 = ∣z1∣2+∣z2∣2|z_1|^2+|z_2|^2 if and only if {z_1}\mathop {{z_2}}\limits^\\_ z is pure imaginary.

Explanation

Solution

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:
∣z1+z2∣2=∣z1∣2+∣z2∣2+2Re⁡∣z1z2∣|{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:
∣z1+z2∣2|z_1+z_2|^2 = ∣z1∣2+∣z2∣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::
∣z1+z2∣2=∣z1∣2+∣z2∣2+2Re⁡∣z1z2∣|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} + 2\operatorname{Re} |{z_1}{z_2}|
We get,
∣z1∣2+∣z2∣2+2Re⁡∣z1z2∣=∣z1∣2+∣z2∣2|{z_1}{|^2} + |{z_2}{|^2} + 2\operatorname{Re} |{z_1}{z_2}| = |{z_1}{|^2} + |{z_2}{|^2}
Cancelling the same terms, gives:
2Re⁡∣z1z2∣=02\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.