Solving z^6=i for arguments between 90 and 180 degrees.

[woof123]

(powers of complex numb.):Find the solution of the following equation whose argument is strictly between 90​ degrees and ​​ 180​ degrees: z^6=i?

I don't understand why the modulus of i is 1 and the argument of i can be 90∘ plus any multiple of 360

 

[HallsofIvy]

Do you know what "modulus" and "argument" are? The "modulus" of a complex number, z= a+ bi, is defined as $$|z|= \sqrt{z\cdot\overline{z}}= \sqrt{a^2+ b^2}$$. If $$z= i$$ then $$\overline{z}= -i$$ so that $$|i|= \sqrt{i\cdot(-i)}= \sqrt{-(i\cdot i)}= \sqrt{-(-1)}= 1$$. The "argument" of a complex number, z= a+ bi, is defined as $$arg(z)= arctan\left(\frac{b}{a}\right)$$. Of course, if z= i, then a= 0 and b= 1 so b/a is not defined. But the tangent function, $$tan(\theta)$$ goes to infinity as $$\theta$$ goes to $$\pi/2$$ so, "by continuity", the argument of i is $$pi/2$$ (your 90 degrees).

Geometrically, if we represent the complex number, z= x+ yi, as a point in the plane, (x, y), then the "modulus" of z is the distance from (x, y) to the origin (0, 0) (just as |x|, with x a real number is the distance on the real-line from x to 0) which is, of course, $$\sqrt{x^2+ y^2}$$. And the argument is the angle that line makes with the positive x-axis. Since the "imaginary axis" (y axis) is perpendicular to the "real axis" (x axis) that angle is 90 degrees.

 

[Prove It]

(powers of complex numb.):Find the solution of the following equation whose argument is strictly between 90​ degrees and ​​ 180​ degrees: z^6=i?

I don't understand why the modulus of i is 1 and the argument of i can be 90∘ plus any multiple of 360

Draw an Argand Diagram. Mark in "i" which is the co-ordinate (0, 1). How far away is it from the origin? This is the modulus...

Can you see it makes a 90 degree angle with the positive real axis? Can you see that if you kept traveling around the circle (so added multiples of 360 degrees) you would get back to the same point?