Calculate [cos(3pi/4) + isin(3pi/4)]*[cos(-5pi/6) + isin(-5pi/6)]

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To calculate (cospi/4+isinpi/4)^3*(cospi/6+isinpi/6)^-5

The De Moivre.s theorem states that (cosx+insinx)^n = consnx+isinnx for all rational n

So  (cospi/4+isinpi/4)^3*(cospi/6+isinpi/6)^-5

= (cospi+isinpi)^(3/4)* (cospi+isinpi)^(-5/6)

= (cos3pi/4+isin3pi/4) (cos(-5pi/6)+isin(-5pi)/6)

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

= sqrt3/4 -i/2sqrt2-isqrt3/4 - 1/4

= (1-sqrt3)/4 - (sqrt2+sqrt3)i/4.

Therefore (cospi/4+isinpi/4)^3*(cospi/6+isinpi/6)^-5 = (1-sqrt3)/4 - (sqrt2+sqrt3)i/4.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply the rule of multiplying 2 complex number, put in polar form:

[cos (a1) + i*sin (a1)]*[cos (a2) + i*sin (a2)] = [cos (a1+a2) + i*sin (a1+a2)]

We'll also apply Moivre's rule:

[cos (a1) + i*sin (a1)]^n = [cos (n*a1) + i*sin (n*a1)]

We'll apply Movre's rule to the first pair of brackets:

[cos(pi/4) + isin(pi/4)]^3 = [cos(3pi/4) + isin(3pi/4)] (1)

We'll apply Movre's rule to the next brackets:

[cos(pi/6) + isin(pi/6)]^-5 = [cos(-5pi/6) + isin(-5pi/6)] (2)

We'll multiply (1) and (2):

[cos(3pi/4) + isin(3pi/4)]*[cos(-5pi/6) + isin(-5pi/6)] = [cos(3pi/4 - 5pi/6) + isin(3pi/4 - 5pi/6)]

[cos(3pi/4 - 5pi/6) + isin(3pi/4 - 5pi/6)] = {cos[(9pi-10pi)/12] + isin[(9pi-10pi)/12]}

[cos(3pi/4 - 5pi/6) + isin(3pi/4 - 5pi/6)] = cos (-pi/12) + isin(-pi/12)

Since cosine function is even and sine function is odd, we'll get:

[cos(3pi/4) + isin(3pi/4)]*[cos(-5pi/6) + isin(-5pi/6)] = cos (pi/12) - isin(pi/12)

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