How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( pi)/6 i)# using trigonometric functions?

Question

How do you evaluate #     e^( ( 7   pi)/4    i)  -   e^(  (    pi)/6      i)# using trigonometric functions?

Carter Anderson · Accepted Answer

#e^{i (7pi)/4} - e^{i pi/6} = (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)#

Use the Euler's Formula, which states that #e^{i theta} -= cos(theta) + i sin(theta)#. (Proof omitted)

Therefore,

#e^{i (7pi)/4} - e^{i pi/6} = (cos((7pi)/4) + i sin((7pi)/4)) - (cos(pi/6) + i sin(pi/6))#

#= (sqrt2/2 + i (-sqrt2/2)) - (sqrt3/2 + i (1/2))#

#= (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)#

Jordan Armstrong · Answer

undefined To evaluate $e^{(7\pi/4)i} - e^{(\pi/6)i}$ using trigonometric functions, first express each complex exponent in trigonometric form.

For $e^{(7\pi/4)i}$:
$$e^{(7\pi/4)i} = e^{(2\pi + \pi/4)i} = e^{\pi/4i}$$

For $e^{(\pi/6)i}$:
$$e^{(\pi/6)i} = e^{(0 + \pi/6)i} = e^{\pi/6i}$$

Now, we can use Euler's formula to express $e^{i\theta}$ in terms of sine and cosine:

$$e^{i\theta} = \cos(\theta) + i\sin(\theta)$$

So, for $e^{\pi/4i}$:
$$e^{\pi/4i} = \cos(\pi/4) + i\sin(\pi/4) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$$

And for $e^{\pi/6i}$:
$$e^{\pi/6i} = \cos(\pi/6) + i\sin(\pi/6) = \frac{\sqrt{3}}{2} + \frac{1}{2}i$$

Substitute these values back into the original expression and simplify:

$$e^{(7\pi/4)i} - e^{(\pi/6)i} = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$$

$$= \frac{\sqrt{2} - \sqrt{3}}{2} + \left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right)i$$