Determine the complex number z if (z'+7i)/z=6?
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Let the complex number z = x + iy. The complement of z , z' = x - iy.
As (z' + 7i)/z = 6
=> ( x - iy + 7i)/ ( x + iy) = 6
=> x - iy + 7i = 6*(x + iy)
=> x - iy + 7i = 6x + 6*i*y
=> x - 6x - iy - 6iy + 7i = 0
=> -5x - 7iy = -7i + 0
Equate the real and complex parts
=> -5x = 0
=> x = 0
-7iy = -7i
=> y = 1
Therefore the complex number z is i.
We can see : (-i + 7i)/i = 6i/i = 6.
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Let z = x+iy. Then z' = x-iy .
So (z'+7i)/z = 6 imlies (x-iy+7i)(x+iy) = 6.
x-iy+7i = 6(x+iy)
x-6x -iy+7i-6iy = 0
-5x + (7-7y)i = 0.
So -5x = 0 and (7-7y) = 0 .
So x = 0. Ot 7-7y = 0. Or 7y= 7, or y = 1.
Therefore z = 0+i = i.
Supposing that z' is the conjugate of z, we'll write z and z' as:
z = a + bi
z' = a - bi
We'll multiply by z both sides:
(z'+7i) = 6z
We'll substitute the expressions of z and z' into the given identity:
(a - bi + 7i) = 6(a+bi)
We'll remove the brackets:
a - bi + 7i = 6a + 6bi
We'll subtract 6a + 6bi:
a - bi + 7i - 6a - 6bi = 0
We'll combine real parts and imaginary parts:
-5a + i(-b + 7 - 6b) = 0
We'll compare and we'll get:
-5a = 0
a = 0
-7b + 7 = 0
7b = 7
b = 1
The complex number z is:
z = 0 + i*1
z = i
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