Let z be a complex number such that |z-6|=5 and |z+2-6 i|=5.

Question

Let $z$ be a complex number such that $|z-6|=5$ and $|z+2-6 i|=5$. Then the value of $z^3+3 z^2-15 z+141$ is equal to :

Accepted Answer

Let $z=x+iy$

$|z-(\alpha +i\beta )|=r$ represents a circle in the complex plane with centre at $(\alpha ,\beta )$ and radius is $r$ unit.

Cartesian equation

$(x-\alpha {)}^{2}+(y-\beta {)}^{2}={r}^{2}$

The equations $|z-6|=5$ and $|z-(-2+6i)|=5$ describe two circles with the same radius $(r=5)$.

Circle 1 : Center ${C}_{1}=(6,0)$

Circle 2 : Center ${C}_{2}=(-2,6)$

Cartesian equation

$(x-6{)}^{2}+{y}^{2}=25\, \, (x+2{)}^{2}+(y-6{)}^{2}=25$

Expanding both:

${x}^{2}-12x+36+{y}^{2}=25\implies {x}^{2}+{y}^{2}-12x+11=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i)$

${x}^{2}+4x+4+{y}^{2}-12y+36=25\implies {x}^{2}+{y}^{2}+4x-12y+15=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(ii)$

Subtracting the first equation from the second to find the common chord:

$\begin{gathered} (4x-(-12x))-12y+(15-11)=0 \ 16x-12y+4=0\implies 4-3y+1=0\implies y=\frac{4x+1}{3} \end{gathered}$

Substitute $y$ back into the first circle equation:

$(x-6{)}^{2}+{(\frac{4x+1}{3})}^{2}=25$

multiply both sides by 9

$\begin{gathered} 9({x}^{2}+36-12x)+16{x}^{2}+8x+1=225 \ 25{x}^{2}-100x+100=0 \ {x}^{2}-4x+4=0 \ (x-2{)}^{2}=0\Rightarrow x=2 \end{gathered}$

put $x=2$ in $y=(4x+1)/3$

$y=(4	imes 2+1)/3=9/3=3$

$z=2+3i$

calculate ${z}^{3}+3{z}^{2}-15z+141$

Compute ${z}^{3}$

$\begin{gathered} {z}^{3}=z\cdot {z}^{2}=(2+3i)(-5+12i) \ =-10+24i-15i+36{i}^{2} \ =-10+9i-36 \ =-46+9i \end{gathered}$

Compute ${z}^{2}$

$(2+3i{)}^{2}=4+12i+9{i}^{2}=4+12i-9=-5+12i$

so

$\begin{gathered} {z}^{3}=-46+9i \ 3{z}^{2}=-15+36i \ -15z=-30-45i \end{gathered}$

Add everything

$(-46+9i)+(-15+36i)+(-30-45i)+141$

Real parts:

$-46-15-30+141=50$

Imaginary parts:

$9i+36i-45i=0$

So ${z}^{3}+3{z}^{2}-15z+141=50$

Solutions · 1