This is the question :
The complex number w has modulus r . Given that w satisfied
500/w = 3|w| + 40i
. Use a non-calculator method to find the value of r. I tried to solve it by calculator and I found that r is approximately equal to 16.2 . I do not know it is correct or not . I hope that I would be given a non-calculator method to solve this question . Thank you for reading!
$\endgroup$3 Answers
$\begingroup$Hint: Take the modulus of both sides, square, now you have a quadratic equation in $q=|w|^2$ that you can solve.
$\endgroup$ 3 $\begingroup$Use the exponential notation: $w=r\,\mathrm e^{i\theta}$. You obtain the equation: \begin{align} &\frac {500}r\mathrm e^{-i\theta}=3r +40 i,\quad\text{whence}\quad \frac{25\cdot10^4}{r^2}=9r^2+1600,\\ \implies &\; 9r^4+1600\,r^2-25\cdot 10^4. \end{align} This is a quadratic equation in $r^2$, with a reduced discriminant $\Delta'=64\cdot 10^4-9\cdot 25\cdot 10^4 =289\cdot 10^4=17^210^4$.
$r^2$ is the positive root: $-800+1700=900$. The rest is straightforward.
$\endgroup$ $\begingroup$another method hint : $$w=a+bi \to \frac 1w=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}+i\frac{-b}{a^2+b^2}$$so $$\frac{500}{w}=3|w|+40i\\ 500(\frac{a}{a^2+b^2}+i\frac{-b}{a^2+b^2})=3\sqrt{a^2+b^2}+40i\\\to \begin{cases}\frac{500a}{a^2+b^2}=3\sqrt{a^2+b^2} & \\\frac{-500b}{a^2+b^2}=40 & \end{cases}$$
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