How can we solve complex equations with roots in the first quadrant?

[Moridin]

Homework Statement

Determine how many roots the equation

$$(z + \frac{i\sqrt{3}}{2})^{29} = \frac{1+i}{\sqrt{2}}$$

has that are in the first quadrant.

The Attempt at a Solution

I would like to treat the right hand side in the following way.

$$(z + \frac{i\sqrt{3}}{2})^{29} = \frac{1+i}{\sqrt{2}}$$

$$z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}$$

It seems reasonable to rewrite the left hand side into

$$z + \frac{i\sqrt{3}}{2} = z + i \sin \frac{\pi}{3}$$

Which give us

$$z + i \sin \frac{\pi}{3} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}$$

Now the real part of the LHS must match the real part of the RHS. This means that the real part of z, must be

$$Re ~z~ = \cos \frac{\pi}{4 \cdot 29}$$

and that the imaginary part of x must be

$$Im ~z ~= \sin \frac{\pi}{4 \cdot 29} - \sin \frac{\pi}{3}$$

From here, I am pretty much lost.

 

[tiny-tim]

Hi Moridin!

Yes, that all looks pretty good (and nice LaTeX, by the way)!

Your only error is in:

$$z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}$$

because (something)^(1/29) has 29 roots.

(much as √(something) = (something)^(1/2) has 2 roots)

So it should be:

$$z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \left(\frac{\pi}{4 \cdot 29}\,+\,\frac{2n \pi}{29}\right) + i \sin \left(\frac{\pi}{4 \cdot 29}\,+\,\frac{2n \pi}{29}\right)$$

for 0 ≤ n ≤ 28.