A complex number z= 3 + 4i is rotated about another fixed Complex number1+ 2i in anticlockwise by 45 degrees .find the new position of z in the argand plane?

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Answer 1 · 2018-04-17T06:10:26

The new position is #=1+i(2+2sqrt2)#

For a rotation of center #Omega# and angle #theta#, we have

#z'-z_Omega=e^(itheta)(z-z_Omega)#

Here,

#z=3+4i#

#z_Omega=1+2i#

#theta=pi/4#

#i^2=-1#

Therefore,

#z'-(1+2i)=e^(ipi/4)((3+4i)-(1+2i))#

#z-z_Omega=3+4i-1-2i=2+2i#

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

#e^(ipi/4)(z-z_Omega)=(sqrt2/2+isqrt2/2)(2+2i)#

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

#=sqrt2(1+2i+i^2)#

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

#=2sqrt2i#

Finally,

#z'-(1+2i)=2sqrt2i#

#z'=2sqrt2i+1+2i#

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