# Write the complex number in trigonometric form (homework question)

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): https://i.sstatic.net/FCthN.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 $θ= -77.32$.

• $\theta = 45^\circ$ 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! May 15, 2015 at 0:47
• For future reference, adding \$dollar signs \$ around your math statements will format it. Add \$\$ double dollars \$\$ to have the statements centered. May 15, 2015 at 0:49
• 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. May 15, 2015 at 0:51
• @Jared So I would simply replace 45 degrees with 135 degrees because theta is in quadrant two? The rest is correct? $√2(cos135+isin135)$ ? May 15, 2015 at 1:00
• I submitted the answer, but it still says that it is wrong. May 15, 2015 at 1:07

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