# 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.

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**