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: Understanding the Concept:
This problem involves an analytic function given in terms of \(x\) and \(y\). We need to find its derivative \(f'(z)\) and express it in terms of \(z\) to identify the constant \(c\). A good strategy is to first express \(f(z)\) in terms of \(z\) and \(\bar{z}\), and then check for analyticity. An easier way is to express \(f(z)\) directly in terms of \(z\) by recognizing combinations of \(x\) and \(y\).

Step 2: Express f(z) in terms of z:
Let \( u(x,y) = x^2-y^2-2xy \) and \( v(x,y) = x^2-y^2+2xy \). We know that \( z^2 = (x+iy)^2 = x^2-y^2+2ixy \). Let's try 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 \(f(z)\). So, \( f(z) = (1+i)z^2 \).

Step 3: Find the Derivative and the Constant c:
Since \(f(z)\) is a polynomial in \(z\), it is analytic everywhere. We can differentiate it directly with respect to \(z\): \[ f'(z) = \frac{d}{dz}((1+i)z^2) = (1+i)(2z) = 2(1+i)z \] We are given that \( f'(z) = cz \). By 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 \( \sqrt{a^2+b^2} \). \[ |c| = |2+2i| = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \]