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Mathematics Question on Complex Numbers and Quadratic Equations

Let z = x + iy be a complex number. The equation arg (z+1z)=π4\left( \frac{z + 1}{z}\right) = \frac{\pi}{4} represents

A

x2+x+y+y2=0x^2 + x + y + y^2 = 0

B

x2−x+y+y2=0x^2 - x + y + y^2 = 0

C

x2+x−y+y2=0x^2 + x - y + y^2 = 0

D

x2+x+y−y2=0x^2 + x + y - y^2 = 0

Answer

x2+x+y+y2=0x^2 + x + y + y^2 = 0

Explanation

Solution

We have, z=x+iyz = x + iy and arg(z+1z)=π4 arg (\frac{z+1}{z} )= \frac{\pi}{4}
∵z+1z=x+iy+1x+iy\because\frac{ z+1}{z} = \frac{x +iy +1}{x+iy}
=((x+1)+iy)(x−iy)x2+y2= \frac{\left(\left(x+1\right)+iy\right)\left(x -iy\right)}{x^{2}+y^{2}}
⇒z+1z=x2+x+y2+(xy−xy−y)ix2+y2\Rightarrow \frac{z+1}{z} = \frac{x^{2}+x+y^{2}+ \left(xy -xy -y\right)i}{x^{2} +y^{2}}
=x2+y2+x−yix2+y2=\frac{ x^{2}+y^{2} +x -yi}{x^{2} + y^{2}}
∴arg(z+1z)=tan−1(−yx2+y2+x)=π4\therefore arg \left(\frac{z+1}{z}\right) = tan^{-1} \left(\frac{-y}{x^{2} +y^{2} +x}\right) = \frac{\pi}{4}
−yx2+y2+x=tanπ4=1-\frac{y}{x^{2}+y^{2} +x} = tan \frac{\pi}{4} = 1
⇒−y=x2+y2+x\Rightarrow -y = x^{2} +y^{2} + x
⇒x2+y2+x+y=0\Rightarrow x^{2} +y^{2} + x +y = 0