Question

simplify [(√2+iota √3)+(√2-√3)]/[(√3+ iota √2)+(√3-iota√2)​

Answer 1

Answer:

Let #z=-1/2+isqrt3/2#

We have to write #z# in trigonometric form in order to apply DeMoivre's theorem

#z=r(costheta+isintheta)#

#r=1#

#costheta=-1/2#

#sintheta=sqrt3/2#

So, #theta# is located in the 2nd quadrant

#theta=2pi/3#

Therefore, #z=cos((2pi)/3)+isin((2pi)/3)#

We need #z^3#

We use DeMoivre 's theorem,

#(costheta+isintheta)^n=cosntheta+isinntheta#

#z^3=(cos((2pi)/3)+isin((2pi)/3))^3#

#=cos(2pi/3*3)+isin(2pi/3*3)#

#=cos2pi+isin2pi#

#=1#

Answer link



Shwetank Mauria

Nov 23, 2016

#(-1/2+sqrt3/2i)^3=1#

Explanation:

According to DeMoivre's theorem, if #a+ib=rcostheta+irsintheta#

then #(a+ib)^n=r^n(cosntheta+isinntheta)#

Writing #-1/2+sqrt3/2i# in trigonometric form

#-1/2+sqrt3/2i=cos((2pi)/3)+isin((2pi)/3)#

Hence #(-1/2+sqrt3/2i)^3=1^3(cos(3xx(2pi)/3)+isin(3xx(2pi)/3))#

= #cos(2pi)+isin(2pi)#

= #1+i0#

= #1#

Answer 2

Step-by-step explanation:

Given : 8(3a−2b)

2 −10(3a−2b)

By talking (3a−2b) in the given question, we get,

(3a−2b)[8(3a−2b)−10]

By taking 2 as common in the second term,

(3−2b)2[4(3a−2b)−5]

So we get,

8(3a−2b)

2 −10(3a−2b)=2(3a−2b)(12a−8b−5)