Question

Is z=cosx+(i)sinx
Show that z^2= cos2x+(i)sin2x

Answer 1

this is probably easier to read:
\[z=\cos \theta+isin \]
show that
\[z ^{2}=\cos2\theta+isin2\theta\]

Answer 2

If we simply square both sides of \[z = \cos \theta + i \sin \theta \] then we see \[z^2 = (\cos \theta + i \sin \theta)^2\]

Answer 3

Which becomes, \[z^2 = (\cos \theta + i \sin \theta)(\cos \theta + i \sin \theta)\]

Answer 4

Then, \[z^2 = \cos^2 \theta + 2i \cos \theta \sin \theta + i^2 \sin \theta \]

Answer 5

Then, \[z^2 = \cos^2 \theta - \sin^2 \theta + 2i \cos \theta \sin \theta \]

Answer 6

Recall from trigonometry that:\[\cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] and \[\sin(2 \theta) = 2 \sin \theta \cos \theta \]

Answer 7

So, now we have \[z^2 = \cos(2 \theta) + i \sin (2 \theta)\] and we are done.

Answer 8

Thank you! Much appreciated!

Answer 9

No problem. Trig to the rescue! :)