Question

If \[z = \sqrt 2 - i\sqrt 2 \] is rotated through an angle \[45^\circ \] in the anti-clockwise direction about the origin, then the coordinates of its new position are [Kerala (Engg.) \[2005\]]
A) \[(2,0)\]
B) \[(\sqrt 2 ,\sqrt 2 )\]
C) \[(\sqrt 2 , - \sqrt 2 )\]
D) \[(\sqrt 2 ,0)\]
E) \[(4,0)\]

Answer 1

Hint: in this question we have to find the new position of point after rotating the given point through\[45^\circ \]in anti-clockwise direction. First find the angle which given point makes with origin by using appropriate formula then rotate it through\[45^\circ \].

Formula Used: Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one
\[\tan (\theta ) = \dfrac{y}{x}\]

Complete step by step solution: Given: A complex equation
Now we have complex equation equal to\[z = \sqrt 2 - i\sqrt 2 \]
Coordinate of above complex number\[(\sqrt 2 , - \sqrt 2 )\]
We know that
\[\tan (\theta ) = \dfrac{y}{x}\]
\[\theta = {\tan ^{ - 1}}\dfrac{y}{x}\]
\[\theta = {\tan ^{ - 1}}\dfrac{{ - \sqrt 2 }}{{\sqrt 2 }}\]
\[\theta = {\tan ^{ - 1}}( - 1)\]
\[\theta = 135^\circ \]
Now rotate the equation through \[45^\circ \] in anti-clockwise direction about the origin
Now \[\theta \] becomes
\[\theta = 180^\circ \]
\[\theta = {\tan ^{ - 1}}\dfrac{y}{x}\]
\[\theta = {\tan ^{ - 1}}\dfrac{0}{{\sqrt 2 }}\]
So \[x = \sqrt 2 \]and \[y = 0\]
Required coordinates is \[(\sqrt 2 ,0)\]


Option ‘D’ is correct

Note: Generally students make a mistake while finding the value tan in quadrant other than first quadrant. Here in this question vector is rotated through \[45^\circ \]in anticlockwise so the value of new angle becomes \[{180^o}\]so value of tan\[{180^o}\]must be calculated by keeping in mind that angle lies in 2nd quadrant. Complex number is a number which is a union of real and imaginary number. Imaginary part is known as iota. Square of iota is equal to negative one.