Relation from \( \mathbb{C} \) to \( \mathbb{R} \) Defined by \( |x|=y \)
📺 Video Explanation
📝 Question
A relation \( \phi \) from \( \mathbb{C} \) to \( \mathbb{R} \) is defined by:
\[ x \,\phi\, y \iff |x|=y \]
Which of the following is correct?
- (a) \((2+3i)\,\phi\,13\)
- (b) \(3\,\phi\,(-3)\)
- (c) \((1+i)\,\phi\,2\)
- (d) \(i\,\phi\,1\)
✅ Solution
Given:
\[ x\,\phi\,y \iff |x|=y \]
So modulus of complex number must be equal to the given real number.
🔹 Option (a): \((2+3i)\,\phi\,13\)
\[ |2+3i|=\sqrt{2^2+3^2}=\sqrt{13} \]
\[ \sqrt{13}\neq13 \]
❌ False
🔹 Option (b): \(3\,\phi\,(-3)\)
\[ |3|=3 \]
But:
\[ 3\neq -3 \]
Also modulus cannot be negative.
❌ False
🔹 Option (c): \((1+i)\,\phi\,2\)
\[ |1+i|=\sqrt{1^2+1^2}=\sqrt{2} \]
\[ \sqrt{2}\neq2 \]
❌ False
🔹 Option (d): \(i\,\phi\,1\)
\[ |i|=1 \]
✔ True
🎯 Final Answer
\[ \boxed{i\,\phi\,1} \]
✔ Correct option: (d)
🚀 Exam Shortcut
- For complex number \(a+bi\), modulus = \(\sqrt{a^2+b^2}\)
- Modulus is always non-negative
- Compare exact value, not approximation