How do you find #z, z^2, z^3, z^4# given #z=sqrt2/2(1+i)#?

Question

3936 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-17T14:22:03

#z=sqrt2/2(1+i)#, #z^2=i#, #z^3=sqrt2/2(-1+i)# and #z^4=-1#

#z=sqrt2/2(1+i)#

= #(sqrt2/2+isqrt2/2)# and in trigonometric form

= #(cos(pi/4)+isin(pi/4))# or in exponential

= #e^(ipi/4)#

Hence #z^2=e^(i(2pi)/4)=e^(ipi/2)=(cos(pi/2)+isin(pi/2))=i#

and #z^3=e^(i(3pi)/4)=(cos((3pi)/4)+isin((3pi)/4))=-sqrt2/2+isqrt2/2#

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

and

#z^4=e^(i(4pi)/4)=e^(ipi)=(cospi+isinpi)=-1#