Why is the argument of a+bi equal to arctan(b/a)?

Think about the point a+bi on the complex plane. Specifically, a is how far along the x (real) axis, and b is how far up the y (imaginary) axis the point is. If you draw a line connecting the origin and the point a+bi then notice that you've constructed a triangle with sides a, b, and sqrt(a^2+b^2). Recall that tan of an angle = opp/adj, applying this to the triangle gives that the angle between the x-axis and the line from the origin is equal to arctan(b/a). This is exactly what the argument of a complex number is, the angle between the x-axis and the line connecting the number and the origin.

MS

Related Further Mathematics A Level answers

All answers ▸

The roots of the equation z^3 + 2z^2 +3z - 4 = 0, are a, b and c . Show that a^2 + b^2 +c^2 = -2


What is the general solution to the equation d2y/dx2 + dy/dx - 2y = -3sinx + cosx (d2y/dx2 signals a second order derivative)


Convert the general complex number z=x+iy to modulus-argument form.


'Find the first derivative, with respect to x, of arctan(1/x) for non-zero real x. Hence show that the value of arctan(x)+arctan(1/x) is constant for all non-zero x, explicitly stating this constant in your final answer.' How do I solve this?