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?z1+z2?=?z1?+?z2? is possible if:

(a) z 2 = 1 z 1

(b) z 2 = 1 z ? 1

(c) arg (z1)=arg (z2)

(d) z 1 = z 2 (d) 

0 2 Views | Posted 2 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago

    This is an Objective Type Questions as classified in NCERT Exemplar

    L e t           z 1 = r 1 ( c o s θ 1 + i s i n θ 1 )     a n d         z 2 = r 2 ( c o s θ 2 + i s i n θ 2 ) Since     |z1+z2|=|z1|+|z2|                                     z 1 + z 2 = r 1 c o s θ 1 + i r 1 s i n θ 1 + r 2 c o s θ 2 + i r 2 s i n θ 2                                   | z 1 + z 2 | = r 1 2 c o s 2 θ 1 + r 2 2 c o s 2 θ 2 + 2 r 1 r 2 c o s θ 1 c o s θ 2 + r 1 2 s i n 2 θ 1 + r 2 2 s i n 2 θ 2 + 2 r 1 r 2 s i n θ 1 s i n θ 2                                                                   = r 1 2 + r 2 2 + 2 r 1 r 2 c o s ( θ 1 − θ 2 ) B u t                 | z 1 + z 2 | = | z 1 | + | z 2 | S o     r 1 2 + r 2 2 + 2 r 1 r 2 c o s ( θ 1 − θ 2 ) = r 1 + r 2 S q u a r i n g     b o t h     s i d e s ,     w e     g e t             r 1 2 + r 2 2 + 2 r 1 r 2 c o s ( θ 1 − θ 2 ) = r 1 2 + r 2 2 + 2 r 1 r 2 ⇒             2 r 1 r 2 − 2 r 1 r 2 c o s ( θ 1 − θ 2 ) = 0 ⇒                                                 1 − c o s ( θ 1 − θ 2 ) = 0                 ⇒ c o s ( θ 1 − θ 2 ) = 1 ⇒                                                                                     θ 1 − θ 2 = 0               ⇒ θ 1 = θ 2 S o ,                                                                               a r g ( z 1 ) = a r g ( z 2 ) H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( c ) .

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