JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Complex Numbers

More JEE Main questions on Complex Numbers

1. If $x^2+x+1=0$, then the value of ${\left(x+\frac{1}{x}\right)}^4+{\left(x^2+\frac{1}{x^2}\right)}^4+{\left(x^3+\frac{1}{x^3}\right)}^4+\dots +{\left(x^{25}+\frac{1}{x^{25}}\right)}^4$ is :
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2. Let z be the complex number satisfying $|\mathrm{z}-5|\le 3$ and having maximum positive principal argument.
   Then $34{\left|\frac{5z-12}{5iz+16}\right|}^2$ is equal to:
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3. Let $\alpha =\frac{-1+i\sqrt{3}}{2}$ and $\beta =\frac{-1-i\sqrt{3}}{2},i=\sqrt{-1}$. If $(7-7\alpha +9\beta {)}^{20}+(9+7\alpha -7\beta {)}^{20}+(-7+9\alpha +7\beta {)}^{20}+(14+7\alpha +7\beta {)}^{20}={\mathrm{m}}^{10}$, then m is
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4. Let $S=\{z\in \mathbb{C}:4z^2+\overline{z}=0\}$. Then ${\sum }_{z\in S}|z{|}^2$ is equal to :
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5. Let $S=\{z:3\le |2z-3(1+i)|\le 7\}$ be a set of complex numbers. Then ${\mathrm{Min}}_{z\in S}\left|\left(z+\frac{1}{2}(5+3i)\right)\right|$ is equal to:
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6. If $\mathrm{z}=\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2},\mathrm{i}=\sqrt{-1}$, then ${\left({\mathrm{z}}^{201}-\mathrm{i}\right)}^8$ is equal to
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