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 May 1st, 2017, 06:29 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Complex Value Hello Everyone! I want to know if the below answer is correct or not ? How many complex numbers $z$ are there such that $|z+1|= |z+i|$ and $|z|=$ 5 ? Options : A) 0 B)1 C)2 D)3 If you calculate the mod of $z+1$ and $z+i$, they will never become equal (because $i^2$ = -1) I guess the answer would be option A. Please correct me if I'm wrong.
 May 1st, 2017, 08:13 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405 $z = a+bi$ $|z| = 5 \implies \sqrt{a^2+b^2} = 5 \implies a^2+b^2 = 25$ $z+1 = a+bi + 1 = (a+1) + bi$ $z+i = a+bi + i = a + (b+1)i$ $|z+1| = |z+i| \implies \sqrt{(a+1)^2 + b^2} = \sqrt{a^2 + (b+1)^2}$ $(a+1)^2 + b^2 = a^2 + (b+1)^2$ $a^2+2a+1+b^2 = a^2+b^2+2b+1$ $2a=2b \implies a=b$ $a^2+b^2 = 25 \implies 2a^2 = 25 \implies a = b = \pm \dfrac{5\sqrt{2}}{2}$ so, how many solutions for $z$ ? Thanks from greg1313, agentredlum and Lalitha183
 May 1st, 2017, 12:15 PM #3 Senior Member   Joined: Aug 2012 Posts: 1,922 Thanks: 534 $\lvert z \rvert = 5$ is a circle of radius $5$ about the origin. The set of points equidistant from $-1$ and $-i$ is a straight line through the origin. How many points of intersection are there between a line through the origin and a circle about the origin? If you draw a picture this is completely obvious. No computation needed. In fact if you note that the line equidistant from $-1$ and $-i$ makes an angle of $\frac{\pi}{4}$ with the $x$-axis, you can read off skeeter's exact solution without any algebra. Thanks from agentredlum Last edited by Maschke; May 1st, 2017 at 12:35 PM.
 May 1st, 2017, 01:04 PM #4 Senior Member   Joined: Aug 2012 Posts: 1,922 Thanks: 534 ps -- The trick to having intuition about a problem like this is to read $\lvert z - c\rvert$ as "the distance from $z$ to $c$". So $\lvert z\rvert = 5$ says that the distance from $z$ to $0$ is $5$. That is, the locus of points satisfying this equation is a circle about the origin with radius $5$. Then, $\lvert z + 1 \rvert$ and $\lvert z + i \rvert$ are the distances from $z$ to $-1$ and $-i$, respectively. What is the locus of points equidistant from two points? It's the perpendicular bisector of the line segment between them. That's a straight line through the origin that splits the difference between the points $-1$ and $-i$. It's just the familiar line defined by $y = x$ in analytic geometry. That line intersects the unit circle at $\pm (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})$, and you scale that solution by $5$ to hit the circle $\lvert z \rvert = 5$. Thanks from agentredlum Last edited by Maschke; May 1st, 2017 at 01:20 PM.
 May 1st, 2017, 05:13 PM #5 Senior Member   Joined: Aug 2012 Posts: 1,922 Thanks: 534 I had some time on my hands ... This is the pic for the unit circle $\lvert z \rvert = 1$. For the problem as stated, scale the circle by a factor of $5$. But the points $-1$ and $-i$ stay where they are. Thanks from agentredlum Last edited by Maschke; May 1st, 2017 at 05:16 PM.
 May 1st, 2017, 11:02 PM #6 Senior Member   Joined: Aug 2012 Posts: 1,922 Thanks: 534 New improved artwork. No more I promise Thanks from agentredlum
 May 2nd, 2017, 01:59 AM #7 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 I support the use of pictures in mathematics.
 May 10th, 2017, 04:13 AM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,198 Thanks: 872 By the way, your statement "If you calculate the mod of z+1 and z+i, they will never become equal" is wrong. When z= 0 the two numbers are 1 and i which both have modulus 1.
May 10th, 2017, 04:30 AM   #9
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Quote:
 Originally Posted by Country Boy By the way, your statement "If you calculate the mod of z+1 and z+i, they will never become equal" is wrong. When z= 0 the two numbers are 1 and i which both have modulus 1.
Well at that time I didn't think about the modulus

 May 17th, 2017, 07:42 AM #10 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 (x+1)$\displaystyle ^{2}$+y$\displaystyle ^{2}$=x$\displaystyle ^{2}$+(y+1)$\displaystyle ^{2}$ x=y x$\displaystyle ^{2}$+y$\displaystyle ^{2}$=25 x=y=$\displaystyle \pm5/\sqrt2$ Now I have a question. From a picture, the minus sign applies. How do you decide that without the picture? EDIT Whoops. Both plus and minus sign apply. It's not just the intersection of the circles. But there is still a question. How do you know both + and - apply? OK There are two points equidistant from two points at a given distance, but it's still geometry. Thanks from Lalitha183 Last edited by zylo; May 17th, 2017 at 07:59 AM.

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