Square root of complex number11/7/2023 A complex number can always be represented in the polar form: R (cos+isin). but I will answer the 'aperiodic or periodic formula' part. I am not giving a full answer, because I don't use Mathematica. But I don't have 50 reputation to do that.I am studying the principal square root function of complex numbers. 1 Answer Sorted by: 4 The ArcTan part is not weird. Running this code: from cmath import sqrt a 0.2 b 0. And in that case, it would be better if I can give the link in the comments for brevity, instead of typing this long answer. Square root of complex numbers in python Ask Question Asked 7 years, 3 months ago Modified 7 years, 3 months ago Viewed 18k times 19 I have run across some confusing behaviour with square roots of complex numbers in python. In the following description, (z) stands for the complex number and (z) for the absolute value. Note : If b is negative, b/b 1, x and y have different signs. Square root formula for a complex number. This video clarifies misconceptions about square roots of negative numbers. Shortcut to Find Square Root of Complex Number - Questions. 'i' is equal to the square root of -1, and every complex number can be expressed as a + bi, where a and b are real numbers. In fact, I think a link of De Moivre's formula is enough to explain the arctan part. The video explores the intriguing concept of imaginary numbers, specifically the imaginary unit 'i'. We perform the derivation in a motivated way. So, it is a result of an aperiodic function times a "looks like periodic" function (in fact, arctan is not periodic), therefore, it is aperiodic. We show how to explicity compute the square roots of a complex number that is given in rectangular form. But it would be nice to find an explicit representation for. We have (R2*(cosα+isinα))^2 = R2^2*(cosα+isinα)^2Īccording to De Moivre's formula, (cosα+isinα)^2 = cos2α+isin2αīecause Z2 is the square root of Z1, now we have R2^2*(cos2α+isin2α) = R1*(cosθ+isinθ) In other words, every complex number has a square root. Suppose your target number (the square root) is Z2 = R2*(cosα+isinα). but I will answer the "aperiodic or periodic formula" part.Ī complex number can always be represented in the polar form: R*(cosθ+isinθ). Is there a way to specify to Mathematica give me only the positive roots at each $x$? Clear this output is periodic and can have -ve values. Now, from basic complex number algebra, we know that the real part of the $\sqrt)$. Since we want to find the square root of the complex number is negative eight, let's give our square root some form and call them equal. Let's say we try to find the square root of the complex number $(8-x +6i)$. The issue arises when we have fractions as indices of complex numbers. This is a pretty simple issue with Mathematica but I could not find any previous discussion on this topic. For finding the square root of a positive real number, we can use the otherwise called, exponent operator as mentioned here, but for negative or complex.
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