# Homework Help: Solve for phase on the real part of a simple complex equation

1. Dec 3, 2011

### atrus_ovis

1. The problem statement, all variables and given/known data

Solve for theta:
Re { (1+j) e} = -1

2. Relevant equations
euler's formula?
e = cosθ + jsinθ

3. The attempt at a solution
(1+j) e =e+ Je = cosθ+jsinθ +j(cosθ+jsinθ)
=cosθ+jsinθ+jcosθ+j2sinθ = cosθ-sinθ+j(cosθ+sinθ)

The real part equals -1 => cosθ-sinθ = -1

I don't think i'm supposed to find an angle that this is true... - Have i taken a wrong turn somewhere?

2. Dec 3, 2011

### Curious3141

Yes, you're supposed to solve that trig equation. Hint: there are two solutions in [0, 2pi)

To solve that equation, start by observing that the LHS can be expressed as $\sqrt{2}\cos(\theta + \frac{\pi}{4})$

3. Dec 3, 2011

### Curious3141

Probably the most direct way to do it is to start with:

$$(1+j) = \sqrt{2}e^{j(\frac{\pi}{4})}$$

so

$$(1+j)e^{j\theta} = \sqrt{2}e^{j(\frac{\pi}{4} + \theta)} = \sqrt{2}(\cos(\frac{\pi}{4} + \theta) + j\sin(\frac{\pi}{4} + \theta))$$

And you can equate the real part of that expression to -1 to get to where you landed up using the longer way.

Last edited: Dec 3, 2011
4. Dec 3, 2011

### atrus_ovis

Thanks, both of you.

@Curious
yeap, should have converted the cartesian complex number into a phasor right away.

Anyway, taking the real part i end up to :
$cos(\frac{\pi}{4} + \theta)= \frac{-1}{\sqrt{2}}$

(If i use the cosine of a sum of angles formula, i end up to what i got at the first post).

So if you add pi/4 = 45 deg to the angle theta, you get an angle whose cosine is -1.
So theta = 180 - 45 = 135 degrees?

Why two values in 0,2pi ?

5. Dec 3, 2011

### Curious3141

"Both of you"? Well, "both of me" welcome you. I posted again (rather than edit my original post) for clarity as I saw you were already online and might have read the old post.

Generally, in complex analysis, you work in radians unless otherwise directed. Certainly you shouldn't "mix" radians and degrees as you did.

So let's do it in radians, then just state the answers in degrees in parantheses for completeness.

You have: $\cos(\frac{\pi}{4} + \theta) = \frac{-1}{\sqrt{2}}$

There are two angles in [0, 2pi) that give a cosine of $\frac{-1}{\sqrt{2}}$, namely $\frac{3\pi}{4}$ (or 135 degrees) and $\frac{5\pi}{4}$ (or 225 degrees). You can see this by recognising that the principal angle that gives the correct absolute value of $\frac{1}{\sqrt{2}}$ (note the positive sign) for the cosine is $\frac{\pi}{4}$ (or 45 degrees). Cosine is positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants, so find the equivalent angle measures in the 2nd and 3rd quadrants, which are the ones previously stated. Sketch the 4 quadrants if you're unsure how we got there.

So $\frac{\pi}{4} + \theta = \frac{3\pi}{4}$ or $\frac{\pi}{4} + \theta =\frac{5\pi}{4}$

Solving each of those independently gets you $\theta = \frac{\pi}{2}$ (90 degrees) or $\pi$ (180 degrees) respectively. Those are the two values in $[0,2\pi)$, and they are both admissible, which you can verify by plugging back into the original equation. In the first case, $e^{j\theta} = j$ and in the second, $e^{j\theta} = -1$. Multiply those by $(1+j)$ and verify that the real part is $-1$ in each case.

6. Dec 4, 2011

### atrus_ovis

Thanks again, one of you :P

I get it now.Somehow in my head i was trying to figure out cos(θ+π/4) = -1 ... go figure.
For the above, θ does have only 1 solution in (0,2pi) , correct?

7. Dec 4, 2011

### Curious3141

Yes, because pi is the only angle with a cosine equal to -1 within that range.

8. Dec 4, 2011

### atrus_ovis

Yep.Thanks again.