Write the complex number in trigonometric form, once using degrees and once using radians. Begin by sketching the graph to help find the argument θ. (Do not use cis form.)

$$−1 + i$$

My work:

I graphed $x = -1$ and $y = 1$

$$z=r= \sqrt{ x^2 + y^2}$$

$$r= \sqrt{2}$$

$$\tan \theta = \frac{Opposite}{Adjacent} $$

$$\tan \theta = \frac{-1}{1} = -1$$

$$\theta= 45^\circ$$

When put into trig form: $$\sqrt{2} (\cos 45^\circ +i \sin 45^\circ)$$

Here is how my submitted answer looks (it is #9): http://i.imgur.com/hrrg6hg.png

I also need help with $9 − 40i$ (instructions: convert the complex number to trigonometric form. (Enter the angle in degrees rounded to two decimal places. Do not use cis form.).

I went through the same steps as I did on the other problem, and I got $r=41$ and $θ= -77.32$.

  • 2
    $\begingroup$ $\theta = 45^\circ$ isn't correct--this would be in the first quadrant, but $-1 + i$ is in the second quadrant...and besides $\arctan(-1) \neq 45^\circ$...p.s., you should never take the arc-trig of a negative value--you always decide the angle based on the quadrant! $\endgroup$ – Jared May 15 '15 at 0:47
  • $\begingroup$ For future reference, adding \$ dollar signs \$ around your math statements will format it. Add \$\$ double dollars \$\$ to have the statements centered. $\endgroup$ – suneater May 15 '15 at 0:49
  • 2
    $\begingroup$ Also, adding on to Jared's comment, look at the sketch you made. Does that point look like it's at $45^\circ$, as measured from the positive x-axis? On such a problem, the sketches are nice. If you find an inconsistency with your sketch, that should be an alarm. $\endgroup$ – suneater May 15 '15 at 0:51
  • $\begingroup$ @Jared So I would simply replace 45 degrees with 135 degrees because theta is in quadrant two? The rest is correct? $√2(cos135+isin135)$ ? $\endgroup$ – redhound2441 May 15 '15 at 1:00
  • $\begingroup$ I submitted the answer, but it still says that it is wrong. $\endgroup$ – redhound2441 May 15 '15 at 1:07

You have that:$x = - 1, y = 1, z = x+iy \Rightarrow \theta = \pi+\tan^{-1}\left(\dfrac{y}{x}\right) = \pi+\tan^{-1}(-1) = \pi+\dfrac{-\pi}{4}=\dfrac{3\pi}{4}, r= \sqrt{x^2+y^2}=\sqrt{2}$


You should bear in mind that an arctan will always output an answer between -90 and 90, so you may have to add or subtract 180 from the answer to get the appropriate angle, because if you examine your graph, the number appears not to be at -45.


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