Question

the complex number z is defined by z = a + ib, where a and b are real. The complex conjugate of z is denoted by z*.

i) Show that IzI^2 = zz* and (z-ki)* = z* + ki

In an Argand diagram a set of points representing complex numbers z is defined by the equation
I z − 10i I = 2 I z − 4i I.

(ii) Show, by squaring both sides, that zz* − 2iz* + 2iz − 12 = 0.

Answer 1

\[\begin{align*}
|z|&=|a+bi|\\
&=\sqrt{a^2+b^2}\\
|z|^2&=a^2+b^2\\\\
zz^*&=(a+bi)(a-bi)\\
&=a^2+abi-abi-b^2i^2\\
&=a^2+b^2
\end{align*}\]
\[\begin{align*}
(z-ki)^*&=(a+bi-ki)^*\\
&=(a+(b-k)i)^*\\
&=a-(b-k)i\\
&=a-bi+ki\\
&=z^*+ki
\end{align*}\]Easy enough?

Answer 2

For the second part, you'll be using these properties.
\[\begin{align*}
|z-10i|&=2|z-4i|\\
|z-10i|^2&=\left(2|z-4i|\right)^2\\
(z-10i)(z-10i)^*&=4(z-4i)(z-4i)^*&\text{acc. to the first property}\\
(z-10i)(z^*+10i)&=4(z-4i)(z^*+4i)&\text{acc. to the second property}
\end{align*}\]
Expand and simplify what terms you can.