For any two complex numbers z1,z2 we have |z1+z2|2=|z1|2+|z2|2. Then
We know that,
|z1+z2|2=|z1|2+|z2|2+2Re(z1¯¯¯z2)…(i)
Given,
|z1+z2|2=|z1|2+|z2|2…(ii)
From equation (i) and (ii), we get
2Re(z1¯¯¯¯¯z2)=0
⇒z1¯¯¯¯¯z2+¯¯¯¯¯¯¯¯¯z1¯¯¯¯¯z2=0 (∵2Re(z)=z+¯¯¯z,¯¯¯¯¯¯z=z)
⇒z1¯¯¯¯¯z2+¯¯¯¯¯z1z2=0
⇒z1¯¯¯¯¯z2=−¯¯¯¯¯z1z2⇒z1z2=−¯¯¯¯¯z1¯¯¯¯¯z2⇒z1z2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=0(∵¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=¯¯¯¯¯z1¯¯¯¯¯z2)
⇒2Re(z1z2)=0
⇒Re(z1z2)=0
Alternate Solution
We know that,
|z1+z2|2=|z1|2+|z2|2+2Re(z1¯¯¯z2)…(i)
Given,
|z1+z2|2=|z1|2+|z2|2…(ii)
From equation (i) and (ii), we get
2Re(z1¯¯¯¯¯z2)=0
⇒Re(z1z2¯¯¯¯¯z2z2)=0
[∵z¯¯¯z=|z|2]
⇒Re(z1z2|z2|2)=0
⇒z1z2|z2|2 is purely imaginary
So, z1z2 will also be purely imaginary
∴Re(z1z2)=0