If 2+i1+2i=r(cosθ+isinθ ) Then

2+i1+2i=r(cosθ+isinθ ) ⟹(2+i)(2−i) (1+2i)(2−i) =r(cosθ+isinθ ) ⟹54+3i=r(cosθ+isinθ ) ...(1) ⟹∣∣54+3i ∣∣=r ∴r=1From eq. (1) r(cosθ+isinθ )=54+3i ⟹tanθ =43∴θ=arctan43 Ans: C

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If {1+2i}{2+i}=r ( +i ) Then

Question

If $\frac{1 + 2 i}{2 + i} = r (cos θ + i sin θ)$ Then

r = 2

θ = t a n^{- 1} (\frac{4}{3})

r = 5

θ = t a n^{- 1} (\frac{3}{4})

r = 1

θ = t a n^{- 1} (\frac{3}{4})

None of these

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\frac{1 + 2 i}{2 + i} = r (cos θ + i sin θ)

Then

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Text solutionVerified

\frac{1 + 2 i}{2 + i} = r (cos θ + i sin θ)

⟹ \frac{( 1 + 2 i ) ( 2 - i )}{( 2 + i ) ( 2 - i )} = r (cos θ + i sin θ)

⟹ \frac{4 + 3 i}{5} = r (cos θ + i sin θ)

...(1)

⟹ ∣ ∣ \frac{4 + 3 i}{5} ∣ ∣ = r

∴ r = 1

From eq. (1)

r (cos θ + i sin θ) = \frac{4 + 3 i}{5}

⟹ tan θ = \frac{3}{4} ∴ θ = arctan \frac{3}{4}

Ans: C

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Question Text	If $\frac{1 + 2 i}{2 + i} = r (cos θ + i sin θ)$ Then
Answer Type	Text solution:1
Upvotes	150

If 2+i1+2i​=r(cosθ+isinθ ) Then

Text solutionVerified

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If $\frac{1 + 2 i}{2 + i} = r (cos θ + i sin θ)$ Then