May 11, 2015 · There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.

Aug 30, 2010 · It is true that there are multiple functions which provide inverses for restrictions of the exponential function; but it is not true that "$1^i=1$ and also $1^i=\mathrm e^{-2\pi}$". At best, this is shorthand for the position that the absence of a 'distinguished' inverse function for $\exp(x)$ entails that there is no meaningful convention for ...

Jan 5, 2018 · i We can compute this by multiplying both numerator and denominator by the conjugate, 1+i, of the denominator

Mar 30, 2020 · We are basically asking that what transformation is required to get back to the Identity transformation whose basis vectors are i ^ (1,0) and j ^ (0,1). Transformation A will do the job, since in this case after applying the A transformation we will reach to the identity transformation. That's why AA-1 = I

The place where you typically see $1 {\color{red}/} 0 = \infty$ is when doing arithmetic in the projective line.

Jul 25, 2018 · Let #\tan^{-1}x=\theta\quad (\forall \ \ -\pi/2\le \theta \le \pi/2)# #\tan\theta=x# #\therefore \sin\theta=\frac{\tan\theta}{\sec \theta}#

Apr 4, 2015 · If z^3-1=0, then we are looking for the cubic roots of unity, i.e. the numbers such that z^3=1. If you're using complex numbers, then every polynomial equation of degree k yields exactly k solution. So, we're expecting to find three cubic roots. De Moivre's theorem uses the fact that we can write any complex number as \\rho e^{i \\theta}= \\rho (\\cos(\\theta)+i\\sin(\\theta)), …

Mar 26, 2018 · How is #e^(ipi)= -1# ? and what is #e^(ipi/4)#? Precalculus. 1 Answer

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Jun 8, 2016 · 12i-1 7+3i +(-8+9i) First, open up the bracket: =7+3i -8+9i Next, combine the numbers and i terms: =color(red)7color(blue)(+3i) color(red)(-8)color(blue)(+9i) =color ...