Math, asked by 07suraj, 23 days ago

Class 11Commerce Mathematics Complex numbers​

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Answered by abhi569
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Question:

If z₁ = r₁(cosθ₁ + isinθ₁) = r1(cosA + isinA)

z₂ = r(cosθ₂ + isinθ₂) = r2(cosB + isinB)

Then prove arg(z₁z₂) = θ₁ + θ₂ = A + B

Answer: for convenience, θ₁ = A, θ₂ = B

=> z₁z₂ = r₁(cosA + isinA) r₂(cosB + isinB)

=> z₁z₂ = r₁r₂(cosA + isinA)(cosB + isinB)

=> z₁z₂ = r₁r₂(cosAcosB + icosAsinB + isinAcosB + i²sinAsinB)

=> z₁z₂ = r₁r₂(cosAcosB + i(cosAsinB + sinAcosB) - sinAsinB)

=> z₁z₂ = r₁r₂[(cosAcosB - sinAsinB) + i(sinAcosB + cosAsinB)]

=> z₁z₂ = r₁r₂[cos(A + B) + isin(A + B)]

Hence, arg is the argument(angle).

arg(z₁z₂) = A + B = θ₁ + θ₂ proved.

Let | z₁z₂ | = | z₁ ||z₂ | be true.

=> | r₁r₂{cos(A + B) + isin(A + B)} | = |r₁(cosA + isinA)| |r₂(cosB + isinB) |, is true

=> r₁r₂|cos(A + B) + isin(A + B)| = r₁r₂|(cosA + isinA)| |(cosB + isinB) |, it true

=> |cos(A + B) + isin(A + B)| = |(cosA + isinA)| |(cosB + isinB) |, it true

=> √cos²(A + B) + sin²(A + B) = √(cos²A + sin²A)√(cos²B + sin²B) , is true

=> √1 = √1 * √1, true

=> 1 = 1 , which is true.

Hence, | z₁z₂ | = | z₁ ||z₂ | is true. Proved.

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