# Thread: Sum of two complex numbers problem

1. ## Sum of two complex numbers problem

Hi

I'm having trouble with the problem below.

Let $z_1 = i$ and $z_2 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$

Plot $z_1 + z_2$ on an Argand diagram and deduce that $\tan\left(\frac{3\pi}{8}\right) = 1+\sqrt{2}$

I plotted them and the tangent of the resulting argument was indeed $1+\sqrt{2}$, but I have no idea how to show that it is equal to $\frac{3\pi}{8}$. I've been staring blankly at it for an hour now, I know I'm going to kick myself when I see how it's done! :P

Stonehambey

2. Originally Posted by Stonehambey
Hi

I'm having trouble with the problem below.

Let $z_1 = i$ and $z_2 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$

Plot $z_1 + z_2$ on an Argand diagram and deduce that $\tan\left(\frac{3\pi}{8}\right) = 1+\sqrt{2}$

I plotted them and the tangent of the resulting argument was indeed $1+\sqrt{2}$, but I have no idea how to show that it is equal to $\frac{3\pi}{8}$. I've been staring blankly at it for an hour now, I know I'm going to kick myself when I see how it's done! :P

Stonehambey

$z_1+z_2=\frac{1}{\sqrt{2}}+\frac{\sqrt{2}+1}{\sqrt {2}}i
$

Plotting it on the argand diagram . Let the angle of inclination be $\theta$ .Note that this angle has to be expressed in radians .

Like you , i got $\tan\theta=1+\sqrt{2}$

Thus , $\theta=tan^{-1}(1+\sqrt{2})$

= 67.5 ( convert it to radian )

$
=67.5\times\frac{\pi}{180}
$
, giving what you want to find .

3. How did you get the 67.5? With a calculator? I just assumed there was a more exact way of doing it, unless I'm missing something

4. Originally Posted by Stonehambey
How did you get the 67.5? With a calculator? I just assumed there was a more exact way of doing it, unless I'm missing something
OK ....

Let O be the point representing z = 0, let A be the point representing $z = Re (z_2) = \frac{1}{\sqrt{2}}$, let B be the point representing $z = z_2$ and let C be the point representing $z = z_1 + z_2$.

Consider the triangles OAB and OBC. Angle OBA $= \frac{\pi}{4}$ (why?). Therefore angle OBC $= \frac{3 \pi}{4}$.

But triangle OBC is isoscles (why?). Therefore angle COB = OCB $= \frac{\pi}{8}$ (why?).

Therefore angle COA $= Arg(z_1 + z_2) = \frac{\pi}{4} + \frac{\pi}{8} = \frac{3 \pi}{8}$. Voila!