Question

If \( f(z) = (x^2-y^2-2xy) + i(x^2-y^2+2xy) \) and \( f'(z)=cz \), where c is a complex constant, then \( |c| \) is equals to:

• A. \( \sqrt{3} \)
• B. \( \sqrt{2} \)
• C. \( 3\sqrt{3} \)
• D. \( 2\sqrt{2} \)

Hint:

When given \(f(z)\) as \(u(x,y)+iv(x,y)\), try to spot combinations that come from powers of \(z\). The terms \(x^2-y^2\) and \(2xy\) are tell-tale signs of \(z^2\). This can be much faster than using the Cauchy-Riemann equations and then integrating to find \(f(z)\).

Answer 1

The Correct Option is D

Step 1: Problem Overview:
This problem deals with an analytic function expressed using \(x\) and \(y\). The objective is to determine its derivative, \(f'(z)\), and represent it using \(z\) to identify the constant \(c\). A possible approach is to express \(f(z)\) in terms of \(z\) and \(\bar{z}\) and then verify analyticity. Alternatively, directly express \(f(z)\) using \(z\) by recognizing combinations of \(x\) and \(y\).

Step 2: Expressing f(z) in terms of z:
Define \( u(x,y) = x^2-y^2-2xy \) and \( v(x,y) = x^2-y^2+2xy \). We know that \( z^2 = (x+iy)^2 = x^2-y^2+2ixy \). The goal is to construct \(f(z)\) from \(z^2\). Consider the complex number \( (1+i) \). \[ (1+i)z^2 = (1+i)(x^2-y^2+2ixy) = (x^2-y^2+2ixy) + i(x^2-y^2+2ixy) \] \[ = (x^2-y^2) + 2ixy + i(x^2-y^2) - 2xy \] \[ = (x^2-y^2-2xy) + i(x^2-y^2+2xy) \] This matches the given \(f(z)\). Therefore, \( f(z) = (1+i)z^2 \).

Step 3: Finding the Derivative and the Constant c:
Since \(f(z)\) is a polynomial in \(z\), it is analytic everywhere. Differentiate directly with respect to \(z\): \[ f'(z) = \frac{d}{dz}((1+i)z^2) = (1+i)(2z) = 2(1+i)z \] Given \( f'(z) = cz \). Comparing the two expressions for \(f'(z)\), we find that the constant \(c\) is: \[ c = 2(1+i) = 2+2i \]
Step 4: Calculate the Magnitude |c|:
The magnitude (or modulus) of a complex number \(a+bi\) is \( \sqrt{a^2+b^2} \). \[ |c| = |2+2i| = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \]