Question 25458: (a)Find the (multiplicative) inverse of the complex number (3-i)^2 in the form a+ib, a,b belongs to the reals(R).
(b) solve the equation (((1+i)^5)x)= ((1-i)^3) in the ring C of complex numbers (also as number of the form a+ib, a,b belong to the reals(R).
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! (a)Find the (multiplicative) inverse of the complex number (3-i)^2 in the form a+ib, a,b belongs to the reals(R).
LET A+ib=1/(3-i)^2
(A+iB)(3-i)^2=1
LHS=(A+iB)((9+i^2-6i)=(A+iB)(9-1-6i)=(A+iB)(8-6i)
=8A-6Ai+8Bi-6Bi^2=(8A+6B)+i(8B-6A)=RHS=1=1+0*i...HENCE EQUATING REAL AND IMAGINARY PARTS,WE GET
8A+6B=1.........I
8B-6A=0...OR.....4B-3A=0.............II
2*EQN.I-3*EQN.II GIVES US
16A+12B-2-12B+9A=0
25A=2
A=2/25
B=3A/4...=3*2/(25*4)=3/50
HENCE 1/(3-i)^2 = (2/25)+(3/50)*i
(b) solve the equation (((1+i)^5)x)= ((1-i)^3) in the ring C of complex numbers (also as number of the form a+ib, a,b belong to the reals(R).
(((1+i)^5)x)= ((1-i)^3)
X=(1-i)^3/(1+i)^5=((1-i)^3)((1-i)^5)/((1-i)^5(1+i)^5)=
(1-i)^8/((1+i)(1-i))^5=(1-i)^8/(1-i^2)^5=(1-i)^8/32
NOW WE HAVE
(1-i)^2=1+i^2-2i=1-1-2i=-2i
(1-i)^8={(1-i)^2}^4=(-2i)^4=(-2^4)(i^2)^2=16(-1)^2=16
HENCE WE GET 16/32=1/2 AS THE ANSWER FOR X
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