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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

Let z=x+iyz=x+iy.

Given:

∣5z+4+i∣=∣5z−3+2i∣|5z+4+i|=|5z-3+2i|

Compute each expression:

5z+4+i=5x+5iy+4+i=(5x+4)+i(5y+1)5z+4+i=5x+5iy+4+i=(5x+4)+i(5y+1) 5z−3+2i=5x+5iy−3+2i=(5x−3)+i(5y+2)5z-3+2i=5x+5iy-3+2i=(5x-3)+i(5y+2)

Square both moduli:

(5x+4)2+(5y+1)2=(5x−3)2+(5y+2)2(5x+4)^2+(5y+1)^2=(5x-3)^2+(5y+2)^2

Expand and simplify:

(25x2+40x+16)+(25y2+10y+1)=(25x2−30x+9)+(25y2+20y+4)40x+10y+17=−30x+20y+1370x−10y+4=035x−5y+2=0\begin{aligned} (25x^2+40x+16)+(25y^2+10y+1) &= (25x^2-30x+9)+(25y^2+20y+4)\\ 40x+10y+17 &= -30x+20y+13\\ 70x-10y+4 &= 0\\ 35x-5y+2 &= 0 \end{aligned}

So the cartesian form is:

35x−5y+2=0ory=7x+25\boxed{35x-5y+2=0}\quad\text{or}\quad\boxed{y=7x+\frac{2}{5}}
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