How do you solve with root of an imaginary number without calculator?

Answer 1

When doing roots of imaginary or complex numbers, it's best to work in polar form.

A little background first.

A complex number is represented by a magnitude and an angle. This comes from Euler's Formula (see related link): eiΘ = cos(Θ) + i sin(Θ) {Θ is in radians}. Note that both eiΘ and [cos(Θ) + i sin(Θ)] have a magnitude of 1, so multiply by the magnitude: AeiΘ = Acos(Θ) + Ai sin(Θ).

Now if you have a number [a + bi], the angle Θ = arctan(b/a), but this will give Θ between -pi/2 and pi/2 (-90° & +90°). So to get the other angles, you need to figure what quadrant the complex number is in. If a is positive, then it is left of the imaginary axis and your angle is fine. If a is negative, then you need to add 180° (pi radians) to the angle. Or you can subtract pi radians as well. This works because the 180° turn is on the same line with the same slope, just pointing in the opposite direction. To get the magnitude A, just do sqrt(a2 + b2). Now this is for general complex numbers. The question asked for imaginary numbers, which a = 0, and Θ will be pi/2 for positive imaginaries (b>0) or -pi/2 for b<0. The magnitude is just the absolute value of b.

Now a root is the same as raising to the reciprocal power. So a square root is the same as raising to the 1/2 power. Also, the Nth root will have N answers. Example, square root of 4 is +2 or -2. To show how to use polar form, I'll work with square root of 4, first. So 4 can be represented by 4*ei*0 And also by 4*ei*2*pi. Any multiple of 2*pi can be added, as that is a full turn around a circle which gets you pointing in the same direction. So ei*0 = e0 = 1. So 4*ei*0 = 4. And 4^(1/2) = 2. But the other way is the interesting one: (4*ei*2*pi)^(1/2) = (4^(1/2))*(ei*2*pi)^(1/2). By rules of exponents the second part becomes (ei*2*pi/2) = (ei*pi) which equals -1 (look at the Euler Formula above, cos(pi) = -1 and siin(pi) = 0. So the second root = -2, just like you'd expect.

For imaginaries, (i)^(1/2) = (ei*pi/2)^(1/2) = (ei*pi/4) = sqrt(2)/2 + i*sqrt(2)/2. To get the second root, add 2*pi and use (ei*5*pi/2). So (ei*5*pi/2)^(1/2) = (ei*5*pi/4) = -sqrt(2)/2 - i*sqrt(2)/2. For any Nth root, you will have N answers, so just add 2*i*pi to the 'angle' in the exponent, to get an equivalent angle pointing in the same direction. While you have to use radians when designating the eiΘ, you can designate polar form as A�Θ, so i = 1�90°. Now it still works with multiplying by 1/2 or 1/3 or whatever root you want. Now you add 360° rather than 2*pi radians. So to do a 4th root of i, you have 1�90°, 1�450°, 1�810°, and 1�1170°. Now divide each angle by 4: 1�22.5°, 1�112.5°, 1�202.5°, and 1�292.5° ; but did you want the ith root [like in imaginary number i]. So this would be raising to the 1/i power, and (ei*pi/2)^(i) = (ei*i*pi/2) = (e-pi/2), which, by the way is a real positive number [about 0.2079]. Since i is not a counting number, I don't think you'd need to add any 2pi radians, to get more answers (they'd all be real, and with negative exponents of e, would just get smaller and smaller).