Question

6.
Find the number of complex numbers satisfying [Z] = Z + 1 + 2i.
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Answer 1

Answer:

Let, z=x+iy where x and y are real numbers.So,|z|=z+1+2i⇒x2+y2−−−−−−√=x+iy+1+2i⇒x2+y2−−−−−−√=(x+1)+i(y+2)On comparing the real and imaginary parts, we have,y+2=0⇒y=−2And,x2+y2−−−−−−√=x+1⇒x2+4−−−−−√=x+1 (put y=−2)Squaring on both side,x2+4=x2+1+2x⇒2x=3⇒x=32Thusz=3/2−2i is the solution of given equation.

Step-by-step explanation: