How to find the Im: z, when #z=((1+i)/(1-i*sqrt(3)))^(-2i)#?

Question

1784 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-30T15:12:32

#z=e^((7pi)/6)(cos(ln(1/2))-isin(ln(1/2)))#

In polar form #1+i# can be written as #sqrt2(cos(pi/4)+isin(pi/4))# or #sqrt2xxe^(ipi/4)#

and #1-isqrt3# can be written as #2(cos(-pi/3)+isin(-pi/3))# or #2xxe^(-ipi/3)#

Hence #z=((1+i)/(1-isqrt3))^(-2i)#

= #((sqrt2xxe^(ipi/4))/(2xxe^(-ipi/3)))^(-2i)#

= #(1/sqrt2xxe^(i(pi/4+pi/3)))^(-2i)#

= #(1/sqrt2)^(-2i)xxe^(-2i^2(7pi)/12)#

= #(e^(ln(1/sqrt2)))^(-2i)xxe^(2*(7pi)/12)#

= #e^((7pi)/6)e^(i(-2ln(1/sqrt2))#

= #e^((7pi)/6)(cos(-2ln(1/sqrt2))+isin(-2ln(1/sqrt2)))#

= #e^((7pi)/6)(cos(2ln(1/sqrt2))-isin(2ln(1/sqrt2)))#

= #e^((7pi)/6)(cos(ln(1/2))-isin(ln(1/2)))#