Separate 3 under root minus 1 by 2 minus under root 1 into real and imaginary part and hence find its modulus?

Question

Separate 3   under root minus 1 by 2 minus under root 1 into real and imaginary part and hence find its modulus?

Answer

Solution:

Separating the given expression into real and imaginary parts:

Let's consider the given expression:

3√

-1

/ 2 - √

1

We know that √

-1

= i (imaginary unit)

Therefore, the given expression becomes:

3i / 2 - 1

To separate this expression into real and imaginary parts, we need to multiply the numerator and denominator by the conjugate of the denominator (2 + i):

(3i / 2 - 1) * (2 + i) / (2 + i)

Simplifying the numerator using distributive property:

(3i * 2 + 3i * i - 2) / (4 + 2i - i - 1)

(6i - 3 + 3i) / (3 + i)

Separating the real and imaginary parts:

Real part = (6i - 3) / (3 + i) = (6 / 10) - (12 / 10)i = -0.6i

Imaginary part = 3 / (3 + i) = (3 / 10) + (3 / 10)i = 0.3 + 0.3i

Finding the modulus:

The modulus of a complex number (a + bi) is given by:

|a + bi| = √(a^2 + b^2)

Therefore, the modulus of the given expression (3√

-1

/ 2 - √

1

) is:

|3i / 2 - 1| = √((-0.6)^2 + (0.3)^2) = √(0.45) = 0.671

Conclusion:

The real and imaginary parts of the given expression are -0.6i and 0.3 + 0.3i respectively. The modulus of the expression is 0.671.

Answer

I think under root -1 in both numerator and denominator cancel out and there is no imaginary part and real part should be 3 by 2

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