Question

plz solve this question fast​

Answer 1

Step-by-step explanation:

(x^4+2xi) - (3x^2 + yi) =(3-5i)+(1+2yi)

(x^4-3x^2) + i(2x - y) = (3+1) + I(2y-5)

on comparing real and imaginary parts

x^4 -3x^2 = 4 and 2x-y = 2y - 5

x^4 - 3x^2 - 4 = 0

(x2)2 - 3x2 -4 = 0

(x2)2 -4x2 +x2 -4 = 0

x2(x2-4) + 1(x2-4( = 0

(x2-4)(x2+1) = 0

x2 = 4 or -1

this gives x = 2, -2, i, -i

value of x is to be real

so x = 2, -2

now

2x-y = 2y - 5

3y = 2x+5

for x = 2

3y = 9

y = 3

for x= -2

3y = 1

y = 1/3

hence values of x and y are

(2,3) and (-2,1/3)

Answer 2

Answer:

x=2,-2,i and -i

and

y = 3, 1/3 ,(5+2i)/3 and (5-2i)/3

Step-by-step explanation:

x^4+2xi -(3x²+yi)= 3- 5i+(1+2yi)

Both side separating real and imaginary numbers:

(x^4-3x²) +(2x-y)i =4+(2y-5)i.......................(1)

Thus x^4-3x²=4

x^4-3x²-4=0

(x²-4)(x²+1)=0

x2=4 or x2= -1

so x=2,-2,i and -i

Also equating the imaginary we get from(1)

2x-y=2y-5

3y=2x+5

Putting the values of x we get

y = 3, 1/3 ,(5+2i)/3 and (5-2i)/3