UPSC NDA I 2024 Mathematics Paper

Let's start by simplifying the given equation iz2+z−z+i=0. Notice that the terms z−z cancel each other out:
iz2+i=0
Next, factor out the common factor i from the equation:
i(z2+1)=0
For this product to be zero, either i=0 or z2+1=0. Since i is the imaginary unit √−1 and cannot be zero, we are left with:
z2+1=0
Subtract 1 from both sides:
z2=−1
Taking the square root of both sides, we get:
z=±i
Now we need to find the value of (|z|+1)2. The magnitude (or modulus) of z is calculated as follows:
|z|=|i|=1
Therefore, we have:
|z|+1=1+1=2
Finally, we square this result:
(|z|+1)2=22=4