Question #116816
Determine the complex number z which satisfies the equations |z + 3i| = |z + 5 − 2i|
and |z − 4i| = |z + 2i| simultaneously.
1
Expert's answer
2020-05-19T08:49:28-0400

We have given that a complex number zz satisfy these two equation

z4i=z+2i(1)z+3i=z+52i(2)|z − 4i| = |z + 2i| \hspace{1cm}(1)\\ |z + 3i| = |z + 5 − 2i|\hspace{1cm}(2)

Let's assume the solution is z=z0=x0+y0z=z_0=x_0+y_0 ,thus equation (1) will be,


z04i=z0+2i    z04i2=z0+2i2    (z04i)(z0+4i)=(z0+2i)(z02i)(z2=zz)    6i(z0z0)=12(z0=x0+y0)    12y0=12    y0=1|z_0-4i|=|z_0+2i|\\ \implies |z_0-4i|^2=|z_0+2i|^2\\ \implies (z_0-4i)(\overline{z_0}+4i)=(z_0+2i)(\overline{z_0}-2i) \hspace{1cm}(\because |z|^2=z\overline{z})\\ \implies 6i(z_0-\overline{z_0})=-12 \hspace{1cm}(\because z_0=x_0+y_0)\\ \implies -12y_0=-12 \implies y_0=1

Hence,

z0=x0+i(3)z_0=x_0+i \hspace{1cm}(3)

And similarly, from equation (2) we get,


z0+3i=z0+52i    (x0+i)+3i2=(x0+i)+52i2(z0=x0+i)    x0+4i2=(x0+5)i2    x02+16=(x0+5)2+1    10x0=10    x0=1|z_0 + 3i| = |z_0+ 5 − 2i|\\ \implies |(x_0+i) + 3i|^2 = |(x_0+i) + 5 − 2i|^2 \hspace{1cm}(\because z_0=x_0+i)\\ \implies |x_0 + 4i|^2 = |(x_0 + 5) − i|^2\\ \implies x_{0}^2 + 16 = (x_0 + 5)^2 + 1\\ \implies 10x_0=-10 \implies x_0=-1

Thus,

z0=1+iz_0=-1+i

Therefore, required complex number

z=z0=1+iz=z_0=-1+i

which satisfy the equation (1)&(2)(1) \: \& (2). Hence we are done.


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