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What is the result for SQRT(-4) * SQRT(-9)?

This was posted on Brilliant, tweeted, and went viral, with thousands of views. Yet, even if it looks very simple (or impossible), you need to know complex numbers to answer this question. I won't post the solution, let's see how our members  answer it, and see if the majority (among the answers) corresponds to the solution.

For other stories that went viral, click here. For mathematical challenges, click here. And here is a very difficult question: does digit 1 appear infinitely many times in the decimals of Pi? Nobody has ever answered that question with a mathematical proof. 

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For the question about the decimals of pi. Am assuming that you mean pi expressed in base 2 digits, composed of the numbers zero and one only, and the numbers going to  infinity. I recall hearing from a colleague that the digit sequence can be predicted for the base 2 case, and it is infinite, so evidently, the numerals zero and one appear infinitely many times. 

This calcul is not posed as it should: the notation with radical SQRT is reserved for real numbers where the square root as an unique value for positive numbers because in real numbers, every positive number has an unique square root. But in complex set C, every non zero number has two "square roots": the square roots of -9 are 3i and -3i and those of -4 are 2i and -2i so the product can be either 6 or -6. The problem should have been written: "in C, if we multiply a square root of -9 by a square root of -4, what could be the result ?"
A lot of people are just saying +6 - shouldn't it be +6 or -6?

Amen, bro.

Maiia Bakhova said:

Hello, 

If you consider complex numbers then both SQRT(-4)  and SQRT(-9) are not single-valued unless it is specified that we're to use only their primary roots. 

SQRT(-4) can be 2i or -2i, and SQRT(-9) can be 3i or -3i. Only the first values in each pair are primary roots.

Hence their product is not single valued as well. It can be 6 or -6.

Look at the reference at Brilliant: it is implied that the answer is only one. Therefore they consider primary roots only, and it must be 6.

MATLAB's sqrt returns only the principal value, which under MATLAB's convention is defined to be

"For negative and complex numbers z = u + i*w, the complex square root sqrt(z) returns

sqrt(r)*(cos(phi/2) + 1i*sin(phi/2))

where r = abs(z) is the radius and phi = angle(z) is the phase angle on the closed interval -pi <= phi <= pi."


Michael Brock said:

According to Matlab, the answer is -6. But you probably already know that, so I'm wondering if you might be posing a more general question. I work with complex numbers (digital filter design).

2i*3i = 6i^2 = -6

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