How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?

Question

How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?

1 Answer

Impact of this question

1511 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-09T19:42:46

$e^{\frac{π i}{4}}$ $e^((pi i)/4)$ = $cos (\frac{π}{4}) + i sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$ $cos (pi/4) + i sin (pi/4)= (sqrt2)/2 +i(sqrt2)/2$

$e^{\frac{7 π i}{6}}$ $e^((7pi i)/6)$ = $cos (7 \frac{π}{6}) + i sin (7 \frac{π}{6}) = - cos (\frac{π}{6}) - i sin (\frac{π}{6})$ $cos (7pi/6)+ isin (7pi/6)= -cos (pi/6)-isin (pi/6)$
$= - \frac{\sqrt{3}}{2} - i \frac{1}{2}$ $=-sqrt3/2 -i1/2$
(In the third quadrant both sin and cos are negative)

Thus, $e^{\frac{π i}{4}}$ $e^((pi i)/4)$ - $e^{\frac{7 π i}{6}}$ $e^((7pi i)/6)$ = $\frac{\sqrt{2} + \sqrt{3}}{2} + i \frac{\sqrt{2} + 1}{2}$ $(sqrt2 +sqrt3)/2 +i(sqrt2 +1)/2$

Science

Math

Humanities

... and beyond

How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate eπ4i−e7π6i e^( ( pi)/4 i) - e^( ( 7 pi)/6 i) using trigonometric functions?

1 Answer

Related questions

Impact of this question

How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?