QJEEM23Ma0b5d: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0b5d Solution: Let w 1 be the point obtained by the rotation of z 1 = 5 + 4 i…

To solve the problem, let's break it down step by step.

Step 1 : Find $w_{1}$

Given $z_{1} = 5 + 4i$ .

When you rotate $z_{1}$ by $90^{\circ}$ anticlockwise about the origin, the real part becomes negative of the imaginary part of $z_{1}$ , and the imaginary part becomes the real part of $z_{1}$ .

Therefore, $w_{1}$ becomes :

$w_{1} = -4 + 5i$

Step 2 : Find $w_{2}$

Given $z_{2} = 3 + 5i$ .

When you rotate $z_{2}$ by $90^{\circ}$ clockwise about the origin, the real part becomes the imaginary part of $z_{2}$ , and the imaginary part becomes negative of the real part of $z_{2}$ .

Therefore, $w_{2}$ becomes :

$w_{2} = 5 - 3i$

Step 3 : Calculate $w_{1} - w_{2}$

$w_{1} - w_{2} = (-4 + 5i) - (5 - 3i)$

$w_{1} - w_{2} = -9 + 8i$

Step 4 : Find the principal argument

To find the principal argument, we need to compute the tangent inverse of the ratio of the imaginary part to the real part.

Argument = $\tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right)$

Argument = $\tan^{-1}\left(\frac{8}{-9}\right)$

Argument = $-\tan^{-1}\left(\frac{8}{9}\right)$

Since it's in the third quadrant, the principal argument is :

Argument = $\pi + (-\tan^{-1}\left(\frac{8}{9}\right))$

Argument = $\pi - \tan^{-1}\left(\frac{8}{9}\right)$

So, the correct option is :

Option C : $\pi-\tan^{-1} \frac{8}{9}$