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# Help With Imaginary Numbers

-1
152
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1. Let z be a complex number such that |z - 12| + |z - 5i| = 13. Find the smallest possible value of |z|.

2. If n is the smallest positive integer for which there exist positive real numbers a and b such that (a + bi)^n = (a - bi)^n, compute b/a.

For the first one, I am confused about which solutions to keep and which to not, since there are many when you take off the absolute values.

Any help is appreciated!

Mar 22, 2019

#1
+28159
+4

For 1. note that $$|z| = |a+ib|=\sqrt{a^2+b^2}$$

For 2.

Mar 22, 2019
#4
+1

Your help is very appreciated! Thank you for taking the time to help.

Guest Mar 23, 2019
#2
+23128
+2

1.
Let z be a complex number such that $$|z - 12| + |z - 5i| = 13$$.
Find the smallest possible value of $$|z|$$.

$$\text{Let z=a+bi } \\ \text{Let |z|=\sqrt{a^2+b^2} }$$

$$\begin{array}{|rcll|} \hline z - 12 &=& a+bi - 12 \\ &=&(a-12) +bi \\ |z-12| &=& \sqrt{(a-12)^2+b^2 } \\ \hline z - 5i &=& a+bi - 5i \\ &=& a + (b-5)i \\ |z - 5i| &=& \sqrt{a^2+(b-5)^2} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{|z - 12| + |z - 5i|} & \mathbf{=}& \mathbf{13} \\\\ \sqrt{(a-12)^2+b^2 } + \sqrt{a^2+(b-5)^2} &=& 13 \\ \sqrt{(a-12)^2+b^2 } &=& 13- \sqrt{a^2+(b-5)^2} \quad | \quad \text{square both sides} \\ (a-12)^2+b^2 &=& \left( 13- \sqrt{a^2+(b-5)^2} \right)^2 \\ (a-12)^2+b^2 &=& 13^2-26\sqrt{a^2+(b-5)^2}+ a^2+(b-5)^2 \\ 26\sqrt{a^2+(b-5)^2} &=& 13^2+ a^2+(b-5)^2-(a-12)^2-b^2 \\ 26\sqrt{a^2+(b-5)^2} &=& 13^2+ a^2+b^2-10b+25-b^2-a^2+24a-12^2 \\ 26\sqrt{a^2+(b-5)^2} &=& 13^2+ -10b+25+24a-12^2 \quad | \quad 13^2-12^2 = 5^2=25 \\ 26\sqrt{a^2+(b-5)^2} &=& 50 -10b +24a \quad | \quad : 2 \\ 13\sqrt{a^2+(b-5)^2} &=& 25 -5b +12a \quad | \quad \text{square both sides} \\ 13^2\left(a^2+(b-5)^2\right) &=& (25-5b+12a)^2 \\ 13^2\left(a^2+(b-5)^2\right) &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ 13^2a^2+13^2(b-5)^2 &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ 13^2a^2+13^2(b^2-10b+25) &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ 13^2a^2+13^2b^2-13^2\cdot 10b+13^2\cdot 25 &=& 25^2+25b^2+12^2a^2-250b+24\cdot 25a-120ab \\ (13^2-12^2)a^2+(13^2-5^2)b^2+13^2\cdot 25 &=& (13^2\cdot 10-250)b+600a-120ab+25^2-13^2\cdot 25 \\ 5^2a^2+12^2b^2&=& 1440b+600a-120ab-3600 \\ (5^2a^2+120ab+12^2b^2) &=& 120(5a+12b)-3600 \\ (5a+12b)^2 &=& 120(5a+12b)-3600 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (5a+12b)^2 &=& 120(5a+12b)-3600 \\ (5a+12b)^2 -120(5a+12b)+3600 &=& 0 \\\\ 5a+12b &=& \dfrac{120\pm \sqrt{120^2-4\cdot 3600}}{2} \\ 5a+12b &=& \dfrac{120\pm 0}{2} \\ 5a+12b &=& \dfrac{120}{2} \\ \mathbf{ 5a+12b } &\mathbf{=}& \mathbf{60} \quad \text{this is a line!!! }\\ \text{or}\quad b&=&5-\dfrac{5}{12}a \\ \hline \end{array}$$

The normal vector of this line $$5a+12b-60=0$$, perpendicular to the line is $$\vec{n}=\dbinom{5}{12}$$

$$\begin{array}{|rcll|} \hline \vec{n_0} &=& \dfrac{1}{\sqrt{5^2+12^2}} \dbinom{5}{12} \\\\ &=& \dfrac{1}{\sqrt{13^2}} \dbinom{5}{12} \\\\ &=& \dfrac{1}{13} \dbinom{5}{12} \\\\ &=& \begin{pmatrix} \dfrac{5}{13}\\ \dfrac{12}{13} \end{pmatrix} \\\\ |z|_{\text{min}} &=& \dbinom{0}{5}\cdot \begin{pmatrix} \dfrac{5}{13}\\ \dfrac{12}{13} \end{pmatrix} \\ &=& 0\cdot\dfrac{5}{13}+5\cdot \dfrac{12}{13} \\\\ & = & \dfrac{5\cdot 12}{13} \\\\ & = & \dfrac{60}{13} \\\\ \mathbf{ |z|_{\text{min}} } & \mathbf{=} & \mathbf{4.61538461538} \\ \hline \end{array}$$

The smallest possible value of |z| is 4.61538461538

Mar 22, 2019
#3
+1

Thank you very much for your help and work, especially the thorough solution!

Guest Mar 23, 2019