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Z1 and z2 are two complex numbers such that |z1| = |z2| and arg(z1) + arg(z2) = π, then show that z1 = -z¯2. - Mathematics

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प्रश्न

z1 and z2 are two complex numbers such that |z1| = |z2| and arg(z1) + arg(z2) = π, then show that z1 = `-barz_2`.

योग

उत्तर

Let z1 = r1(cosθ1 + isinθ1)

And z2 = r2(cosθ2 + isinθ2) are polar form of two complex numbers z1 and z2.

Given that: |z1| = |z2

⇒ r1 = r2   ......(i)

And arg(z1) + arg(z2) = π

⇒ θ1 + θ2 = π

⇒ θ1 = π – θ2

Now z1 = r1[cos(π – θ2) + isin(π – θ2)]

⇒ z1 = r1[–cosθ2 + isinθ2]

⇒ z1 = –r1(cosθ2 – isinθ2)   ......(i)

z2 = r2[cosθ2 + isinθ2]

`barz_2` = r1[cosθ2 – isinθ2]   ......[∵ r1 = r2]  .....(ii)

From equation (i) and (ii) we get,

z1 = `-barz_2`.

Hence proved.

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अध्याय 5: Complex Numbers and Quadratic Equations - Exercise [पृष्ठ ९२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 5 Complex Numbers and Quadratic Equations
Exercise | Q 16 | पृष्ठ ९२

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