1
$\begingroup$

A complex number is such that the absolute value of z = the absolute value of (z-3i)

(a) Show that the imaginary part of z is 2/3.

I tried squaring both sides and then I find by equating both sides, that when z = a + bi, b satisfies the equation:

b^2 -7b +9 =0 however this does not give me b = 2/3.

$\endgroup$
0
$\begingroup$

$$|z|=|z-3i|$$

Let$$z_1=a+ib$$

$$z_2=z-3i=a+i(b-3)$$ By comparing Modulus of $z_1$ and $z_2$ $$|z_1|=|z_2|$$ $$a^2+b^2=a^2+(b-3)^2$$ $$a^2+b^2=a^2+b^2-6b+9$$

$$6b=9$$

$$b=\frac32$$

$\endgroup$
  • $\begingroup$ it is |z|=|z-3i| $\endgroup$ – user2250537 Nov 9 '14 at 11:08
  • $\begingroup$ but how did u go from 2nd to 3de line $\endgroup$ – user2250537 Nov 9 '14 at 11:09
  • $\begingroup$ did you square both sides? or how did u go from 2nd to third line $\endgroup$ – user2250537 Nov 9 '14 at 11:12
  • $\begingroup$ wouldnt z^2 = (x+yi)^2=x^2 + 2xyi - y^2 ??? $\endgroup$ – user2250537 Nov 9 '14 at 11:16
1
$\begingroup$

You have $$a^2+b^2=a^2+(b-3)^2\iff6b=9\iff2b=3$$

or $$b^2-(b-3)^2=0\iff\{b-(b-3)\}\{b+(b-3)\}=0$$

$\endgroup$
0
$\begingroup$

Geometry! (In the complex plane...)

The distance $|z|=|z-0|$ between the points $z$ and $0$ and the distance $|z-3\mathrm i|$ between the points $z$ and $3\mathrm i$ coincide if and only if the point $z$ is on the line orthogonal to the segment $[0,3\mathrm i]$ and passing through its middle point $\frac12(0+3\mathrm i)=\frac32\mathrm i$. This is the horizontal line $\frac32\mathrm i+\mathbb R$ of equation $\Im(z)=\frac32$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.