Complex Numbers

I am having a tough time solving this problem: Determine the complex number “z” so that “w”:(z-1-i)/(z+1+i) is a real number. I would aprecciate the help.

#Calculus

Note by Marcelo M
2 years, 8 months ago

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Comments

The answer maybe Infinite Complex Numbers. Here is my solution :

Let us assume the complex number Z=x+iyZ = x + iy.

Then w=Z−1−iZ+1+i=(x−1)+i(y−1)(x+1)+i(y+1)w = \dfrac{Z - 1 - i}{Z + 1 + i} = \dfrac{(x - 1) + i(y - 1)}{(x + 1) + i(y + 1)}

Now, we will rationalize ww. For that we must multiply and divide with (x−1)−i(y−1)(x - 1) - i(y - 1)

  ⟹  w=(x−1)+i(y−1)(x+1)+i(y+1)×(x−1)−i(y−1)(x+1)−i(y+1)\implies w = \dfrac{(x - 1) + i(y - 1)}{(x + 1) + i(y + 1)} \times \dfrac{(x - 1) - i(y - 1)}{(x + 1) - i(y + 1)}

On doing normal multiplication, we will get

w=(x2−1)−i2(y2−1)+i[(y−1)(x+1)−(x−1)(y+1)](x+1)2−i2(y+1)2w = \dfrac{(x^2 - 1) - i^2 (y^2 - 1) + i[(y - 1)(x + 1) - (x - 1)(y + 1)]}{(x + 1)^2 - i^2 (y + 1)^2}

As you know i2=−1i^2 = -1 we can further simplify it as

w=x2−1+y2−1+i[xy+y−x−1−xy−x+y+1]x2+1+2x+y2+1+2y=(x2+y2−2)+i(2y−2x)x2+y2+2(x+y+1)w = \dfrac{x^2 - 1 + y^2 - 1 + i[xy + y - x - 1 - xy - x + y + 1]}{x^2 + 1 + 2x + y^2 + 1 + 2y} = \dfrac{(x^2 + y^2 - 2) + i(2y - 2x)}{x^2 + y^2 + 2(x + y + 1)}

  ⟹  w=x2+y2−2x2+y2+2(x+y+1)+i2y−2xx2+y2+2(x+y+1)\implies w = {\color{#E81990} \dfrac{x^2 + y^2 - 2}{x^2 + y^2 + 2(x + y + 1)}} + i{\color{#3D99F6} \dfrac{2y - 2x}{x^2 + y^2 + 2(x + y + 1)}}

If ww is a real number then the imaginary part is absent, that means it must be equal to 00.

2y−2xx2+y2+2(x+y+1)=0\color{#3D99F6} \dfrac{2y - 2x}{x^2 + y^2 + 2(x + y + 1)} = 0

  ⟹  2y−2x=0  ⟹  y−x=0\implies 2y - 2x = 0 \implies y - x = 0

  ⟹  x=y\implies \large \boxed{\color{#20A900} x = y}

The obtained condition says that if any complex number have both its real part and imaginary part same then it will satisfy the above condition.

Ram Mohith - 2 years, 8 months ago

What is the answer ?

I got it as infinite complex numbers

Ram Mohith - 2 years, 8 months ago

If you want to practice more questions on complex numbers or any other maths topics you can visit my set Mathematics Dine Right. If you have still any more doubts you can ask it here.

Ram Mohith - 2 years, 8 months ago
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