JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 7)

The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there $$2\sqrt 2 $$ units in the south-westwardsdirection. Then its new position in the Argand plane is at the point represented by :

2 + 2i

1 + i

$$-$$1 $$-$$ i

$$-$$2 $$-$$2i

Explanation

Here,

z $$-$$ (3 + 3i) = $$2\sqrt 2 $$ (cos($$-$$135^o) + i sin ($$-$$ 135^o))

= $$2\sqrt 2 $$ ($$-$$ $${1 \over {\sqrt 2 }}$$ $$-$$$${i \over {\sqrt 2 }}$$)

= $$-$$ 2 $$-$$ 2i

$$ \Rightarrow $$ z = 3 + 3 i $$-$$ 2 $$-$$ 2 i = 1 + i

Note :

Polar form of a complex number :

z = r (cos$$\theta $$ + i sin$$\theta $$)

Here r = modulus of z and $$\theta $$ argument of z.

JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 7)

Explanation

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