(3+i)100=299(p+iq)
2100(cos6π+isin6π)100=299(p+iq)
2100ei350π=299(p+iq)
2(cos350π+isin350π)=p+iq
2(cos(17π−3π)+isin(17π−3π))=p+iq
2(−cos3π+isin3π)=p+iq
2(2−1+2i3)=p+iq
p=-1&q=\sqrt{3}
p and q Roots of the equation
x2−(3−1)x−3=0
JEE Main 2021 — Mathematics Algebra
If (3+i)100=299(p+iq), then p and q are roots of the equation :
Held on 26 Aug 2021 · Verified 6 Jul 2026.
x2−(3+1)x+3=0
x2+(3+1)x+3=0
x2+(3−1)x−3=0
x2−(3−1)x−3=0
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