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 ▸

Two planes have eqns r.(3i – 4j + 2k) = 5 and r = λ (2i + j + 5k) + μ(i – j – 2k), where λ and μ are scalar parameters. Find the acute angle between the planes, giving your answer to the nearest degree.


How do you plot a complex number in an Argand diagram?


By forming and solving a suitable quadratic equation, find the solutions of the equation: 3cos(2A)-5cos(A)+2=0


Differentiate arcsin(2x) using the fact that 2x=sin(y)