Question

Find the complex number z satisfying the equations |(z-12)/(z-8i)|=5/3 and |(z-4)/(z-8)|=1 ​

Answer 1

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:\left|\dfrac{z-12}{z-8i}\right|=\dfrac{5}{3}$

$\:\:\:\:\bullet\:\:\:\:\left|\dfrac{z-4}{z-8}\right|=1$

To Find:

• Value of complex number z.

Solution:

$\text{Let, the complex number z be a + ib}\\\\:\implies z=a+ib\\\\\text{We are given that,}\\\\:\implies\left|\dfrac{z-12}{z-8i}\right|=\dfrac{5}{3}\\\\\text{So,}\\\\:\implies\left|\dfrac{(a+ib)-12}{(a+ib)-8i}\right|=\dfrac{5}{3}\\\\:\implies\left|\dfrac{a-12+ib}{a+ib-8i}\right|=\dfrac{5}{3}\\\\\text{Separating real and imaginary parts}\\\\:\implies\left|\dfrac{(a-12)+ib}{a+(b-8)i}\right|=\dfrac{5}{3}\\\\\text{On cross-multiplying,}$

$:\implies3|(a-12)+ib|=5|a+(b-8)i|\\\\:\implies|3(a-12)+3ib|=|5a+5(b-8)i|\\\\\text{We know that |z|=u^2+v^2 (where z=u+iv). So,}\\\\:\implies3^2(a-12)^2+3^2(b^2)=5^2(a^2)+5^2(b-8)^2\\\\:\implies9(a-12)^2+9b^2=25a^2+25(b-8)^2- - - -(1)$

$\text{We are also given that,}\\\\:\implies\left|\dfrac{z-4}{z-8}\right|=1\\\\\text{So,}\\\\:\implies\left|\dfrac{(a+ib)-4}{(a+ib)-8}\right|=1\\\\\text{Separating real and imaginary parts}\\\\\:\implies\left|\dfrac{(a-4)+ib}{(a-8)+ib}\right|=1\\\\\text{On cross-multiplying,}\\\\:\implies|(a-4)+ib|=|(a-8)+ib|\\\\\text{We know that |z|=u^2+v^2 (where z=u+iv). So,}\\\\:\implies(a-4)^2+b^2\!\!\!\!\!/\,=(a-8)^2+b^2\!\!\!\!\!/\\\\:\implies(a-4)^2=(a-8)^2\\\\\text{We know that, (x-y)^2=x^2-2xy+y^2 . So,}$

$:\implies a^2\!\!\!\!\!/\,-8a+16=a^2\!\!\!\!\!/\,-16a+64\\\\:\implies16a-8a=64-16\\\\:\implies8a=48\\\\:\implies\underline{a=6\:}- - - -(2)\\\\\text{Putting value of a from(2) in (1),}\\\\:\implies9(a-12)^2+9b^2=25a^2+25(b-8)^2\\\\:\implies9(6-12)^2+9b^2=25(6)^2+25(b-8)^2\\\\:\implies9(6)^2+9b^2=25(36)+25(b-8)^2\\\\:\implies9b^2-25(b-8)^2=25(36)-9(36)\\\\:\implies9b^2-25(b^2-16b+64)=36(25-9)\\\\:\implies9b^2-25b^2-25(-16b)-25(+64)=36(16)\\\\:\implies-16b^2+400b-1600=16(36)$

$\text{Taking -16 common in LHS,}\\\\:\implies-16\!\!\!\!/\:(b^2-25b+100)=16\!\!\!\!/\:(36)\\\\:\implies b^2-25b+100=-36\\\\:\implies b^2-25b+136=0\\\\\text{On splitting the middle term,}\\\\:\implies b^2-17b-8b+136=0\\\\:\implies b(b-17)-8(b-17)=0\\\\:\implies(b-17)(b-8)=0\\\\:\implies\underline{b = 17\,\,or\,\,8}\\\\\text{So,}\\\\:\implies z=a+ib\\\\:\implies z=(6)+i(17)\,\,\,or\,\,\,z=(6)+i(8)\\\\\bf{:\implies z=6+17i\:\:or\:\:z=6+8i}$

Formulae Used:

• |z| = a² - b²
• (a - b)² = a² - 2ab + b²