If z 1 and z 2 are two complex number and a - b are two real numbers- then - az 1- bz 2-2- bz 1- az 2-2 equalsA- a2-b2-z1-2-z2-2B- 2ab -z1 z2-C- a2-b2-z1 z2-D- a2-b2z12-z22

The correct option is D |az_1 bz_2|^2 + |bz_1 + az_2|^2 = a^2|z_1|^2 + b^2|z_2|^2 ab z_1 z_2 ab z_1 z_2 + b^2 |z_1|^2 + a^2|z_2|^2 + ba z_1 z_2 + ba z ...

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Byju's Answer

Standard XI

Mathematics

Properties of Cube Root of a Complex Number

If z 1 and z ...

Question

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

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Solution

The correct option is B

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

= $a^{2} | z_{1} |^{2} + b^{2} | z_{2} |^{2} - a b z_{1}_{2} - a b_{1} z_{2} + b^{2} | z_{1} |^{2} + a^{2} | z_{2} |^{2} + b a z_{1}_{2} + b a_{1} z_{2}$

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

we have

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. If this is true enter 1, else enter 0.

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If z1 and z2 are two complex number and a,b are two real numbers, then |az1−bz2|2+|bz1+az2|2 equals

The correct option is B |az1−bz2|2+|bz1+az2|2 = a2|z1|2+b2|z2|2−abz1¯¯¯¯¯z2−ab¯¯¯¯¯z1z2+b2|z1|2+a2|z2|2+baz1¯¯¯¯¯z2+ba¯¯¯¯¯z1z2 = (a2+b2)(|z1|2+|z2|2) we have

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

The correct option is B

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

= $a^{2} | z_{1} |^{2} + b^{2} | z_{2} |^{2} - a b z_{1}_{2} - a b_{1} z_{2} + b^{2} | z_{1} |^{2} + a^{2} | z_{2} |^{2} + b a z_{1}_{2} + b a_{1} z_{2}$

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

we have