Let p,q€R and (1-sqrt(3)i)^200= 2^199(p+iq) .Then p+q+q^2 and p-q+q^2 are roots of the equation

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Let p,q€R and(1-i√3)^200=2^199(p+iq). The p+q+q^2 and p-q+q^2 are roots of the equationLet p,q€R and(1-i√3)^200=2^199(p+iq). The p+q+q^2 and p-q+q^2 are roots of the equation
Let \(p, q \in R\) and \((1-\sqrt{3} i)^{200}=2^{199}(p+i q), i=\sqrt{-1}\) Then \(p+q+q^2\) and....Let \(p, q \in R\) and \((1-\sqrt{3} i)^{200}=2^{199}(p+i q), i=\sqrt{-1}\) Then \(p+q+q^2\) and....
Let p,qɛR and (1-sqrt3i)^200=2^199(p+iq),i=(sqrt-1). Then p+q+q^2 and p-q+q^2 are roots of....Let p,qɛR and (1-sqrt3i)^200=2^199(p+iq),i=(sqrt-1). Then p+q+q^2 and p-q+q^2 are roots of....
let p,q belongs R and (1+√3i)^200=2^199(p+iq). Then p+q+q^2 and p-q+q^2 are roots of the equationlet p,q belongs R and (1+√3i)^200=2^199(p+iq). Then p+q+q^2 and p-q+q^2 are roots of the equation
#Let p,q are real number & (1 - root3i)^200 = 2^199(p+iq) then p+q+q^2 & p-q+q^2 are roots of #jee#Let p,q are real number & (1 - root3i)^200 = 2^199(p+iq) then p+q+q^2 & p-q+q^2 are roots of #jee
If (1+3i)²⁰⁰ = 2¹⁹⁹(p+iq) then p+q+q² and p-q+q² are the roots of the equation.. | jee mains 2023If (1+3i)²⁰⁰ = 2¹⁹⁹(p+iq) then p+q+q² and p-q+q² are the roots of the equation.. | jee mains 2023
If (sqrt(3)+i)^(100)=2^(99)(p+iq) then p and q are roots of the equation ______If (sqrt(3)+i)^(100)=2^(99)(p+iq) then p and q are roots of the equation ______
Here’s how You Complete The square #viral #algebraHere’s how You Complete The square #viral #algebra