How do you multiply #e^(( pi )/ 4 i) * e^( 3 pi/2 i ) # in trigonometric form?

Question

1432 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-02T07:04:09

The answer is #=sqrt2/2(1-i)#

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

#i^2=-1#

Therefore,

#e^(pi/4i)=cos(pi/4)+isin(pi/4)=sqrt2/2+isqrt2/2=sqrt2/2(1+i)#

#e^(3/2pii)=cos(3/2pi)+isin(3/2pi)=0-i#

So,

#z=e^(pi/4i)*e^(3/2pii)=sqrt2/2(1+i)*(-i)=sqrt2/2(-i-i^2)#

#=sqrt2/2(1-i)#

#"Verification"#

#z=(cosphi+isinphi)=sqrt2/2(1-i)#

#cosphi=sqrt2/2#

#sinphi=-sqrt2/2#

#phi=-pi/4#, #[2pi]#

#z=e^(pi/4i)*e^(3/2pii)=e^((pi/4+3/2pi)i)=e^(7/4pii)=e^(-1/4pii)#