Why is log i 0.682i Where i is the imaginary number sqr rt -1?

Answer 1

By Euler's formula, e^ix = cosx + i*sinx

Taking natural logarithms, ix = ln(cosx + i*sinx)

When x = pi/2, i*pi/2 = ln(i)

But ln(i) = log(i)/log(e) where log represents logarithms to base 10.

That is, i*pi/2 = log(i)/log(e)

And therefore log(i) = i*pi/2*log(e) = i*0.682188 or 0.682*i to three decimal places.

Answer 2

So your (log) is a log base 10. First let's look at the Natural Log of i. Then we can use a conversion to convert that to log base 10. If we use the notation A*e^(i*θ) notation for complex numbers, where e is the base of natural logarithms, and θ is an angle (in radians) measured from the positive real axis (counterclockwise), and A is the magnitude (straight line distance from the origin). So i is magnitude 1, positioned 90° (pi/2 radians). So now we represent i as e^(i*pi/2). Take the natural log of this, and we have ln(e^(i*pi/2)). We can take the power down, in front of the logarithm and have (i*pi/2)*ln(e), and ln(e) is just 1, so we have ln(i) = (i*pi/2).

But we want log (base 10). To get log (base 10) of x, you can do ln(x)/(ln(10)). So pi/2 = 1.5708, and ln(10) = 2.30259. Divide these and you get 0.68219, but remember it was multiplied by i.

So Log[base 10] of i = i*0.68219