
In the Argand plane, the vector \[z = 4 - 3i\;\] is turned in the clockwise sense through \[{180^o}\] and stretched three times. The complex number represented by the new vector is [DCE2005]
A) \[12 + 9i\]
B) \[12 - 9i\]
C) \[ - 12 - 9i\]
D) \[ - 12 + 9i\]
Answer
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Hint: In this question we have to find the complex number after rotating a given vector through \[{180^o}\]and stretching it three times. First, write the complex number of vectors obtained after rotating the given vector then stretch it 3 times to get the required complex number.
Formula used: Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represents the real part of a complex number
iy is an imaginary part of a complex number
i is iota
The Square of iota is equal to the negative of one
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
Complete step by step solution: Given: A complex equation
Now we have complex equation equal to \[z = 4 - 3i\;\]
Coordinate of above complex number\[(4, - 3i)\]
This coordinate lies in 4th quadrant
Now after rotating vector in clockwise direction through angle \[{180^o}\]we get new vector new vector lies in 2nd quadrant
Let new vector be \[{z_1}\]
\[{z_1} = {e^{ - i\pi }}z = (cos\pi - isin\pi )\]
\[{z_1} = - 4 + 3i\;\]
Modulus of given vector z
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\left| z \right| = \left| {{z_1}} \right| = \sqrt {{4^2} + {{( - 3)}^2}} = 5\]
The unit vector in direction of new vector
\[\dfrac{{\overrightarrow {{z_1}} }}{{\left| {{z_1}} \right|}} = \dfrac{{ - 4}}{5} + \dfrac{3}{5}i\]
Now stretched the vector 3 times
\[3\left| {{z_1}} \right|(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i)\]
\[3\left| {{z_1}} \right|(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i) = 15(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i)\]
\[15(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i) = - 12 + 9i\]
The required complex number is \[ - 12 + 9i\]
Thus, Option (D) is correct.
Note: Generally students make a mistake while finding the value sin or cos in a quadrant other than the first quadrant. Here in this question vector is rotated through \[{180^o}\] so the value of sin\[{180^o}\] and cos\[{180^o}\]must be calculated by keeping in mind that these are lies in 2nd quadrant. A complex number is a number that is a union of real and imaginary numbers. The imaginary part is known as an iota. The Square of iota is equal to the negative one.
Formula used: Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represents the real part of a complex number
iy is an imaginary part of a complex number
i is iota
The Square of iota is equal to the negative of one
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
Complete step by step solution: Given: A complex equation
Now we have complex equation equal to \[z = 4 - 3i\;\]
Coordinate of above complex number\[(4, - 3i)\]
This coordinate lies in 4th quadrant
Now after rotating vector in clockwise direction through angle \[{180^o}\]we get new vector new vector lies in 2nd quadrant
Let new vector be \[{z_1}\]
\[{z_1} = {e^{ - i\pi }}z = (cos\pi - isin\pi )\]
\[{z_1} = - 4 + 3i\;\]
Modulus of given vector z
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\left| z \right| = \left| {{z_1}} \right| = \sqrt {{4^2} + {{( - 3)}^2}} = 5\]
The unit vector in direction of new vector
\[\dfrac{{\overrightarrow {{z_1}} }}{{\left| {{z_1}} \right|}} = \dfrac{{ - 4}}{5} + \dfrac{3}{5}i\]
Now stretched the vector 3 times
\[3\left| {{z_1}} \right|(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i)\]
\[3\left| {{z_1}} \right|(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i) = 15(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i)\]
\[15(\dfrac{{ - 4}}{5} + \dfrac{3}{5}i) = - 12 + 9i\]
The required complex number is \[ - 12 + 9i\]
Thus, Option (D) is correct.
Note: Generally students make a mistake while finding the value sin or cos in a quadrant other than the first quadrant. Here in this question vector is rotated through \[{180^o}\] so the value of sin\[{180^o}\] and cos\[{180^o}\]must be calculated by keeping in mind that these are lies in 2nd quadrant. A complex number is a number that is a union of real and imaginary numbers. The imaginary part is known as an iota. The Square of iota is equal to the negative one.
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