Answer:
Answer b of the first one
answer of q9
Step-by-step explanation:
(x+iy)(2−3i)=4+i
2x−(3x)i+(2y)i−3yi
2
=4+i
Real
2x+3y
+
Imaginary
(2y−3x)
i=4+i
Comparing the real & imaginary parts,
2x+3y=4--------------------------(1)
2y−3x=1----------------------------(2)
Solving eq(1) & eq(2),
4x+6y=8
−9x+6y=3
13x=5⇒x=
13
5
y=
13
14
∴(x,y)=(
13
5
,
13
14
)
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SIMILAR QUESTIONS
star-struck
If e
x+iy
=α+iβ, then x+iy is called logarithm of α+iβ to the base e.∴log
e
(x+iy)=log
e
(re
iθ
) =log
e
r+iθ where r is modulus value of x+iy & θ be the argument of x+iy If i
α+iβ)
=α+iβ, then α
2
+β
2
equals
Hard
View solution
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The modulus of (1 + i) (1 + 2i) (1 + 3i) is equal to
