Math question on Complex numbers. Demovrie's Theorem involved.?

I assume that you are supposed to find the sin and cos and divide them.

The rule is (r cis theta)^n = r^n cis (n theta). (You missed the second n.) We can use that to write (cis 40)^3 = cis 120. But cis 120 has the exact form cis 120 = -1/2 + sqrt(3)i/2. So

cis 40 = (-1/2 + sqrt(3)i/2)^(1/3)

cos 40 + i sin 40 = (-1/2 + sqrt(3)i/2)^(1/3)

Now we can use the real and imaginary part functions to separate the sin and cos.

cos 40 = Re[(-1/2 + sqrt(3)i/2)^(1/3)]

sin 40 = Im[(-1/2 + sqrt(3)i/2)^(1/3)]

Finally, since tan 40 is sin 40 / cos 40,

tan 40 = Im[(-1/2 + sqrt(3)i/2)^(1/3)] / Re[(-1/2 + sqrt(3)i/2)^(1/3)]

A technicality needs to be pointed out. The 1/3 power, or cube root, can be multivalued in complex analysis. Every complex number except 0 can have three cube roots. But there is a so-called principal value, and if it is understood that that's what we mean here, then the cube root of cis 120 will be cis 40.

uncooked D is right....right that's yet in a various thank you to look at it. z^2= - 4i so we are looking forward to that z is a complicated extensive form interior the form a + bi so if z = a +bi z^2 = (a+bi)^2 = a^2 + 2ab i +b^2 i^2 considering i^2 = -a million we are able to declare that z^2 = a^2 +2ab i - b^2 which equals 0 - 4 i comparing genuine and imaginary factors of a^2 + 2ab i - b^2 = 0 - 4 i we see genuine: a^2-b^2=0 , so a^2 = b^2, and a = b or a = -b (considering the two squared turns into beneficial) IMAGINARY: 2ab i = -4i, so 2ab = - 4, and ab = -2. From genuine steps above: of course a = - b no longer + b or ab does no longer be detrimental. considering a = -b we are able to sub in to the ab = -2 equation with a = -b so if ab = -2 and a = -b then - b^2 = - 2, b ^2 = 2, and b = +sqrt 2 or -sqrt 2 considering a equals -b we see while b = pos. sqrt 2 then a = neg sqrt 2 and while b = neg sqrt 2 then a = pos sqrt 2. considering z=a+bi, subbing those 2 opportunities for a and b yields the two available solutions: z = ( V2 - i * V2 ) OR z = ( - V2 + i * V2 )