For any two complex numbers z 1 z 2 and any two real numbers a and b - a z 1- b z 2-2- bz 1- az 2-2-A- a-b-z1-2-z2-2B- -a2-b2- z1-2-z2-2C- -a2-b2- z1- z2-D- None of these

The correct option is B |a nbsp;z_1 nbsp; b nbsp;z_2 nbsp;|^2 + |bz_1 + az_2 |^2 nbsp; = (az_1 bz_2)(a z1 b z2 ) + (a z_1 + nbsp;b z_2)(a z1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Algebra of Complex Numbers

For any two c...

Question

For any two complex numbers $z_{1} z_{2}$ and any two real numbers a and b $| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$ =

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

z_{1}

z_{2}

{| (a z_{1} - b z_{2}) |}_{1}^{2} + {| b z_{1} + a z_{2} |}_{1}^{2} =

z_{1}, z_{2}

a

b

| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2} =

z_{1}, z_{2}

a, b

| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2} = (a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2})

If $z_{1} a n d z_{2}$ are two complex number and a,b are two real numbers, then

$| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$ equals

z_{1}, z_{2}

{| a z_{1} - b z_{2} |}_{1}^{2} + {| b z_{1} + a z_{2} |}_{1}^{2} = (a^{2} + b^{2}) ({| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2})

Join BYJU'S Learning Program