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To find the angle $\theta$ of this complex number I know that I have to imagine it in the complex graph, draw a triangle and then calculate the arctan.

Here is the representation of $-3+5i$ on the graph:

enter image description here

(I'll explain $\alpha$ and $\phi$ in a moment)

I know that since I have the adjacent and opposite sides of the triangle I can correlate them with the angle using the tangent. Then, I need to calculate the arctan to get the angle.

But my problem is finding the tangent, because I don't know which is the right triangle in this case.

My question is: is $tan \theta = \frac{5}{-3}$ or $tan \theta = \frac{-3}{5}$?

How do I know whether if $\theta$ is supposed to be $\alpha$ or $\phi$?

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  • $\begingroup$ $\theta$ is always the angle that the complex number makes with the positive part of the real axis, in counterclockwise direction. So in this case, the angle would, by just immediately guessing, be somewhere in the vincinity of $120^\circ = \frac{2\pi}3$. So it's neither $\alpha$ not $\phi$. $\endgroup$
    – Arthur
    Dec 14, 2016 at 18:37
  • $\begingroup$ @Arthur So that would be $\alpha + 90$, in degrees... $\endgroup$
    – Segmentation fault
    Dec 14, 2016 at 18:38
  • $\begingroup$ Yes, that's exactly what it would be. $\endgroup$
    – Arthur
    Dec 14, 2016 at 18:38
  • $\begingroup$ It's always arctan of Im(z)/Re(z) as im(x) is always r*sin (a) and re(z) is always r*cos(a). So it's a=arctan (5/-3) = 180 - theta = 90 + alpha. [arctan (-b) = 180 - arctan b = 90 + arctan (1/b) if restricted to first and second quadrant as your pictures demonstrate.] $\endgroup$
    – fleablood
    Dec 14, 2016 at 19:12
  • $\begingroup$ @TheGreatDuck I don't know about the unit circle but I know polar coordinates... The tangent is the slope of the line? Does that mean that if I had a regular graph and drew a circle with it's center on where a linear equation crosses either the x or y axis and calculated the tangent using that circle it would be correct? I had never thought about it that way. Interesting $\endgroup$
    – Segmentation fault
    Dec 14, 2016 at 19:14

3 Answers 3

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By convention your $\theta$ is defined to be

the angle made with the positive $X$-axis in anticlockwise direction

In this case it will be $90^\circ+\alpha$

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Angle of complex number is angle the complex vector makes with positive part of real (usually $x$) axis in counterslockwise direction. So, the answer is $\alpha+\frac\pi2$.

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Remember that lengths are positive.

The green triangle with angle $\varphi$ has an adjacent of length $3$ and an opposite of length $5$.

Hence $\tan \varphi = \frac{5}{3}$ and so $\varphi = \arctan \frac{5}{3} \approx 1.03$.

The red triangle with angle $\alpha$ has an adjacent of length $5$ and an opposite of length $3$.

Hence $\tan \alpha = \frac{3}{5}$ and so $\alpha = \arctan \frac{3}{5} \approx 0.54$.

Notice that $\varphi + \alpha = \arctan \frac{5}{3} + \arctan \frac{3}{5} = \frac{1}{2}\pi$, as expected.

Now, the argument of a complex number is by definition the angle made with the positive real axis. So to find the argument of $-3+5\mathrm i$ you stand at the origin looking at the positive reals, e.g. at number $1 = 1+0\mathrm i$. To face the number $-3+5\mathrm i$ you need to turn anti-clockwise to the positive imaginary axis, and then continue on through the angle $\alpha$.

Hence, $\arg(-3+5\mathrm i) = \frac{1}{2}\pi + \alpha = \frac{1}{2}\pi+\arctan\frac{3}{5} \approx 2.11$

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