complex number question.

JEE Mains Guidance › Algebra questions › complex number question.

x+iy=(1-i*root3)^100.........x=?y=?

#Mathematics #Algebra

UP 0 DOWN 0 0 2
Post on FacebookPost anonymously?

2 Answers

1708

man111 singh ·Jul 31 '11 at 22:49

$\hspace{-16}\mathbf{x+iy=\left(1-i\sqrt{3}\right)^{100}=(-2)^{100}.\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)^{100}}$\\\\\\ As We Know that $\mathbf{-\frac{1}{2}+i\frac{\sqrt{3}}{2}=\omega}\;\;$(Cube Root of Unity.)\\\\\\ So $\mathbf{x+iy=2^{100}.\omega^{100}=2^{100}.\omega}$\\\\\\ So $\mathbf{x+iy=2^{100}.\left(-\frac{1}{2}+\frac{i.\sqrt{3}}{2}\right)=-2^{99}+i.2^{99}.\sqrt{3}}$\\\\\\ so $\boxed{\boxed{\mathbf{\left(x,y\right)=\left(-2^{99}\;, 2^{99}.\sqrt{3}\right)}}}$$

UP 0 DOWN 0
Post on FacebookPost anonymously?

3

h4hemang ·Jul 31 '11 at 23:01

thanks friend.
i have just started complex numbers so i don't know that much.

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω