Express tan inverse (X +iY) in the form of (A+iB).And Express sin inverse (X +iY) in the form of (A+iB).

To express the inverse tangent and inverse sine of a complex number in the form of \( A + iB \), where \( A \) and \( B \) are real numbers, we can use specific formulas that relate to complex analysis. Let's break this down step by step for both functions.

Inverse Tangent of a Complex Number

The formula for the inverse tangent of a complex number \( z = X + iY \) is given by:

tan^-1(z) = (i/2) * ln((1 - iz) / (1 + iz))

Substituting \( z = X + iY \) into the formula, we have:

tan^-1(X + iY) = (i/2) * ln((1 - i(X + iY)) / (1 + i(X + iY)))

Now, simplifying the expressions inside the logarithm:

• Calculate \( 1 - i(X + iY) = 1 - iX + Y \)
• Calculate \( 1 + i(X + iY) = 1 + iX - Y \)

Thus, we can rewrite the expression as:

tan^-1(X + iY) = (i/2) * ln(((1 - iX + Y) / (1 + iX - Y)))

To express this in the form \( A + iB \), we need to evaluate the logarithm. The logarithm of a complex number can be expressed as:

ln(r) + iθ

where \( r \) is the modulus and \( θ \) is the argument of the complex number. After calculating these values, we can separate the real and imaginary parts to find \( A \) and \( B \).

Inverse Sine of a Complex Number

For the inverse sine function, the formula for a complex number \( z = X + iY \) is:

sin^-1(z) = -i * ln(iz + sqrt(1 - z²))

Substituting \( z = X + iY \), we have:

sin^-1(X + iY) = -i * ln(i(X + iY) + sqrt(1 - (X + iY)²))

Next, we need to compute \( 1 - (X + iY)² \):

• Calculate \( (X + iY)² = X² - Y² + 2iXY \)
• Thus, \( 1 - (X + iY)² = 1 - (X² - Y² + 2iXY) \)

This gives us:

1 - X² + Y² - 2iXY

Now, we can find the square root of this expression, which will also be a complex number. After substituting back into the logarithm, we can express \( sin^-1(X + iY) \) in the form \( A + iB \) by separating the real and imaginary parts.

Final Thoughts

Both of these transformations involve complex logarithms and square roots, which can be intricate. However, by carefully following the steps and using the properties of logarithms and complex numbers, you can derive the desired forms for both the inverse tangent and sine functions. If you have specific values for \( X \) and \( Y \), we could work through those calculations together for a clearer understanding!