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The line joining the origin and the point represented by z=1+i is rotated through an angle \frac{\pi}{2} in an anticlockwise direction about the origin and stretched by additional \sqrt {3} units. The new position of the point is

Option: 1

\\\mathrm{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(-1+i)}


Option: 2

\\\mathrm{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(1-i)}


Option: 3

\\\mathrm{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(-1-i)}


Option: 4

\\\mathrm{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}}(1+i)}


Answers (1)

best_answer

As we have leart,

Rotation Theorem (Coni Method)

Three points A, B and C in the argand plane whose affixes are z1, z2 and z3 respectively.

Let suppose we want to rotate AB to AC.

\\\mathrm{\frac{z_3-z_1}{z_2-z_1}=\frac{ \left | z_3-z_1 \right |}{\left | z_2-z_1 \right |}e^{i\theta}}

Note: Final vector should be in numberator and starting vector in denominator. \theta is positive if rotation is anti-clockwise and negative if it is clockwise.

 

Now,

Let initial point be A (1 + i), O be origin (0+0i), and B(z) be the final point

|AO| = \sqrt{2}

And length of OB is \sqrt{3} more than |AO|, so |BO| = \sqrt{2} + \sqrt{3}

Now using Rotation Theorem

\\\frac{OB}{OA} = \frac{|OB|}{|OA|}.e^{\frac{i\pi}{2}}\\\\ \frac{z-0}{1+i-0} = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}}e^{\frac{i\pi}{2}}\\\\ z = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}}.(1+i).e^{\frac{i\pi}{2}}\\\\ z = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}}.(1+i).(cos(\pi/2) + isin(\pi/2))

\\ z = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}}.(1+i).i\\\\ z = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}}.(-1+i)

Correct option is (a)

Posted by

Nehul

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