True or False?

Level 1

If $i = \sqrt{-1}$ . We can say that $2i > i$

False True

2 solutions

Aniket Verma
Mar 5, 2015

This question is simply based on the fact that imaginary numbers can't be compared.

Vincent Cagampang
Mar 6, 2015

Since 2i^2<i^2

Sorry to say this sir but this is not a perfect answer to the questions.

Purushottam Abhisheikh - 6 years, 3 months ago

I disagree and assert that 2i > i is true. Let i = exp[i * (4m+1) pi/2] = exp[i * (4n+1) pi/2], where m & n are integers (not necessarily equal). This gives:

2 exp[i * (4m+1) pi/2] > exp[i * (4n+1)*pi/2],

or 2 > exp[ (i*pi/2) * ((4n+1) - (4m+1))],

or 2 > exp[ (i*pi/2) * 4(n-m)],

or 2 > exp[i * (n-m) * 2pi] = 1 (for all integers m & n)

The RHS of the above inequality contains an integer multiple of 2pi radians, which exp[i * 2k * pi] = 1 for ALL integers k.

tom engelsman - 3 years, 10 months ago

True or False?

2 solutions

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