Question

find the real value of x and y for which:
(1-i)x+(1+i)y=1-3i

Answer 1

(1-i)x+(1+i)y=1-3i
(x+y)-(x-y)i=1-3i
so...................
x+y=1.............................1
x-y=3.....................................2
adding both eq
2x=4
x=2
so
y=-1

Answer 2

The value of x = 2 & y = - 1

Given :

$\displaystyle \sf (1-i)x+(1+i)y=1-3i$

To find :

The value of x and y

Concept :

Complex Number

A complex number z = a + ib is defined as an ordered pair of Real numbers ( a, b) that satisfies the following conditions :

(i) Condition for equality :

(a, b) = (c, d) if and only if a = c, b = d

(ii) Definition of addition :

(a, b) + (c, d) = (a+c, b+ d)

(iii) Definition of multiplication :

(a, b). (c, d) = (ac-bd , ad+bc )

Of the ordered pair (a, b) the first component a is called Real part of z and the second component b is called Imaginary part of z

Solution :

Step 1 of 3 :

Write down the given equation

Here the given equation is

$\displaystyle \sf (1-i)x+(1+i)y=1-3i$

Step 2 of 3 :

Form the equations to find the value of x and y

$\displaystyle \sf (1-i)x+(1+i)y=1-3i$

$\displaystyle \sf{ \implies }(x + y) + ( - x + y)i = 1 - 3i$

By the property of equality we get

x + y = 1 - - - - - - (1)

- x + y = - 3 - - - - - (2)

Step 3 of 3 :

Find the value of x and y

Adding Equation 1 and Equation 2 we get

2y = - 2

$\displaystyle \sf{ \implies }y = \frac{ - 2}{2}$

$\displaystyle \sf{ \implies }y = - 1$

Putting y = - 1 in Equation 1 we get

x - 1 = 1

$\displaystyle \sf{ \implies }x = 1 + 1$

$\displaystyle \sf{ \implies }x = 2$

Hence the required value of x = 2 & y = - 1

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