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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 101 of 150

Marks: +1, -0

z=\frac{13-5 i}{4-9 i}, i=\sqrt{-1}

Solution:

Concept:

For a quotient of complex numbers,

\arg\left(\frac{z_1}{z_2}\right)=\arg z_1 - \arg z_2

Use standard tangent subtraction to simplify the result.

Explanation:

Let

z=\frac{13-5i}{4-9i}

Then

\arg z = \arg(13-5i) - \arg(4-9i)

Both

13-5i

and

4-9i

lie in the fourth quadrant, so their principal arguments are negative.

\arg(13-5i)=-\tan^{-1}\left(\frac{5}{13}\right)

\arg(4-9i)=-\tan^{-1}\left(\frac{9}{4}\right)

Therefore,

\arg z=-\tan^{-1}\left(\frac{5}{13}\right)+\tan^{-1}\left(\frac{9}{4}\right)

Using

\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)

\arg z=\tan^{-1}\left(\frac{\frac{9}{4}-\frac{5}{13}}{1+\frac{9}{4}\cdot\frac{5}{13}}\right)

Compute the numerator and denominator:

\frac{9}{4}-\frac{5}{13}=\frac{117-20}{52}=\frac{97}{52}

1+\frac{9}{4}\cdot\frac{5}{13}=1+\frac{45}{52}=\frac{97}{52}

So the fraction equals

1

, giving

\arg z=\tan^{-1}(1)

Since

\tan\left(\frac{\pi}{4}\right)=1

, we get

\arg z=\frac{\pi}{4}

Answer:

\frac{\pi}{4}

, which is option A.

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