If z1 and z2 are two complex numbers and a, b are two real numbers, then |a z1-b z2|2+|b z1+a z2|2 equals

Question

If z1 and z2 are two complex numbers and a, b are two real numbers, then |a z1-b z2|2+|b z1+a z2|2 equals (A) (a2+b2)|z1 z2| (B) (a2+b2)(z12+z22) (C) (a2+b2)(|z1|2+|z2|2.) (D) 2 a b|z1 z2|

Tardigrade Admin · Accepted Answer

We have |a z1-b z2|2+|b z1+a z2|2 = a2|z1|2+b2|z2|2-a b z1 barz2-a b barz1 z2 +b2|z1|2+a2|z2|2+b a z barz2+b a barz1 z2 = (a2+b2)(|z1|2+|z2|2)

Q. If $z_{1}$ and $z_{2}$ are two complex numbers and $a, b$ are two real numbers, then $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2}$ equals

$(a^{2} + b^{2}) ∣ z_{1} z_{2} ∣$

$(a^{2} + b^{2}) (z_{1}^{2} + z_{2}^{2})$

$(a^{2} + b^{2}) (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$

$2 ab ∣ z_{1} z_{2} ∣$

Q. If z1​ and z2​ are two complex numbers and a,b are two real numbers, then ∣az1​−bz2​∣2+∣bz1​+az2​∣2 equals

(a2+b2)∣z1​z2​∣

(a2+b2)(z12​+z22​)

(a2+b2)(∣z1​∣2+∣z2​∣2)

2ab∣z1​z2​∣

We have ∣az1​−bz2​∣2+∣bz1​+az2​∣2 =a2∣z1​∣2+b2∣z2​∣2−abz1​zˉ2​−abzˉ1​z2​+b2∣z1​∣2+a2∣z2​∣2+bazzˉ2​+bazˉ1​z2​ =(a2+b2)(∣z1​∣2+∣z2​∣2)

Q. If $z_{1}$ and $z_{2}$ are two complex numbers and $a, b$ are two real numbers, then $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2}$ equals

$(a^{2} + b^{2}) ∣ z_{1} z_{2} ∣$

$(a^{2} + b^{2}) (z_{1}^{2} + z_{2}^{2})$

$(a^{2} + b^{2}) (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$

$2 ab ∣ z_{1} z_{2} ∣$