若(1+3i)四次方=x+yi,则x= ,y=

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若(1+3i)四次方=x+yi,则x= ,y=
若(1+3i)四次方=x+yi,则x= ,y=

若(1+3i)四次方=x+yi,则x= ,y=
首先知道i^2=-1
(1+3i)^2=1^2+2*1*3i+(3i)^2
=1+6i+9i^2
=1+6i-9
=6i-8
(6i-8)^2=(6i)^2-2*6i*8+8^2
=36i^2-96i+64
=-36-96i+64
= 28-96i
所以(1+3i)^4=[(1+3i)^2]^2
=(6i-8)^2
=28-96i
=x+yi,
所以x=28,y=-96

X=28,Y=-96