3dB Frequency of an LED Transfer Function

Homework Statement

According to my textbook:

The LED transfer function ##H(\omega_m)## is defined as:

$$H(\omega_m) = \frac{1}{1+j\omega_{m}\tau_{c}}$$

The 3-dB modulation bandwidth ##f_{\text{3 dB}}## is defined as the modulation frequency at which ##H(\omega_m)## is reduced by 3 dB or by a factor of 2. The result is:

$$f_{\text{3 dB}}=\sqrt{3}(2\pi\tau_{c})^{-1} \tag{1}$$

I don't understand how they derived the last expression.

The Attempt at a Solution

Let ##G(j\omega_{m})=\frac{1}{1+j\omega_{m}\tau_{c}}##.

##|G(j\omega_{m})|^{2}=\frac{1}{1+j\omega_{m}\tau_{c}}.\frac{1}{1-j\omega_{m}\tau_{c}}=\frac{1}{1+\omega_{m}^{2}\tau_{c}^{2}}##

##\therefore |G(j\omega_{m})|=\frac{1}{\sqrt{1+\omega_{m}^{2}\tau_{c}^{2}}}##

And ##|G(0)|=1##, therefore to find the 3 dB point we must solve:

$$\frac{|G(j\omega)|}{|G(0)|}=\frac{1}{\sqrt{1+\omega_{\text{3 dB}}^{2}\tau_{c}^{2}}}=\frac{1}{\sqrt{2}}$$

However when I solve this I get:

$$\omega_{\text{3 dB}}=\frac{1}{\tau_{c}}\ \text{or}\ f_{\text{3 dB}}=\frac{1}{2\pi\tau_{c}}$$

So where does the ##\sqrt{3}## factor in equation (1) come from? What is the mistake here?

Any help would be greatly appreciated.