If z₁=5-2 i and z₂=3+i, where i={-1}, then ≤ft({z₁+z₂}{z₁-z₂}) is

Correct Answer is : tan⁡−1(2219) ^{-1}≤ft(22/19)tan−1(1922)

Correct Answer is : tan⁡−1(2219) ^{-1}≤ft(22/19)tan−1(1922)

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MHTCET 16 May 2024 Shift 2 Paper

Question : 104 of 150

Marks: +1, -0

z_{1}=5-2 i

z_{2}=3+i

i=\sqrt{-1}

\arg \left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)

Solution:

Concept:

For a complex number

z = a + ib

with

a > 0

, the argument is given by

\arg(z) = \tan^{-1}\left(\frac{b}{a}\right)

Explanation:

First, calculate

z_1 + z_2

and

z_1 - z_2

z_1 + z_2 = (5 - 2i) + (3 + i) = 8 - i

z_1 - z_2 = (5 - 2i) - (3 + i) = 2 - 3i

Therefore,

\frac{z_1 + z_2}{z_1 - z_2} = \frac{8 - i}{2 - 3i}

Rationalize the denominator by multiplying the numerator and denominator by the conjugate

2 + 3i

\frac{8 - i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} = \frac{(8 - i)(2 + 3i)}{2^2 + 3^2}

Expand the numerator:

(8 - i)(2 + 3i) = 16 + 24i - 2i - 3i^2 = 16 + 22i + 3 = 19 + 22i

So,

\frac{z_1 + z_2}{z_1 - z_2} = \frac{19 + 22i}{13} = \frac{19}{13} + \frac{22}{13}i

The real part is

\frac{19}{13}

and the imaginary part is

\frac{22}{13}

Thus,

\arg\left(\frac{z_1 + z_2}{z_1 - z_2}\right) = \tan^{-1}\left(\frac{\frac{22}{13}}{\frac{19}{13}}\right) = \tan^{-1}\left(\frac{22}{19}\right)

Answer:

Option A:

\tan^{-1}\left(\frac{22}{19}\right)

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