JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 17)
Let z0 be a root of the quadratic equation, x2 + x + 1 = 0, If z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$, then arg z is equal to :
$${\pi \over 4}$$
$${\pi \over 6}$$
$${\pi \over 3}$$
0
Explanation
1 + x + x2 = 0
x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$
z0 = w, w2
Now
z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$
z = 3 + 6iw81 $$-$$ 3iw93 (w93 = w81 = 1)
$$ \Rightarrow $$ z = 3 + 3i
then arg(z) = tan$$-$$1$$\left( {{3 \over 3}} \right)$$ = tan$$-$$1 (1) = $${\pi \over 4}$$
x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$
z0 = w, w2
Now
z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$
z = 3 + 6iw81 $$-$$ 3iw93 (w93 = w81 = 1)
$$ \Rightarrow $$ z = 3 + 3i
then arg(z) = tan$$-$$1$$\left( {{3 \over 3}} \right)$$ = tan$$-$$1 (1) = $${\pi \over 4}$$
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