May 1st, 2017, 06:29 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 221 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: 2,724 Thanks: 1378 
$z = a+bi$ $z = 5 \implies \sqrt{a^2+b^2} = 5 \implies a^2+b^2 = 25$ $z+1 = a+bi + 1 = (a+1) + bi$ $z+i = a+bi + i = a + (b+1)i$ $z+1 = z+i \implies \sqrt{(a+1)^2 + b^2} = \sqrt{a^2 + (b+1)^2}$ $(a+1)^2 + b^2 = a^2 + (b+1)^2$ $a^2+2a+1+b^2 = a^2+b^2+2b+1$ $2a=2b \implies a=b$ $a^2+b^2 = 25 \implies 2a^2 = 25 \implies a = b = \pm \dfrac{5\sqrt{2}}{2}$ so, how many solutions for $z$ ? 
May 1st, 2017, 12:15 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,780 Thanks: 482 
$\lvert z \rvert = 5$ 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. Last edited by Maschke; May 1st, 2017 at 12:35 PM. 
May 1st, 2017, 01:04 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,780 Thanks: 482 
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 $\lvert z \rvert = 5$. Last edited by Maschke; May 1st, 2017 at 01:20 PM. 
May 1st, 2017, 05:13 PM  #5 
Senior Member Joined: Aug 2012 Posts: 1,780 Thanks: 482 
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. Last edited by Maschke; May 1st, 2017 at 05:16 PM. 
May 1st, 2017, 11:02 PM  #6 
Senior Member Joined: Aug 2012 Posts: 1,780 Thanks: 482 
New improved artwork. No more I promise 
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: 2,970 Thanks: 807 
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 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 221 Thanks: 2  
May 17th, 2017, 07:42 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 
(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. Last edited by zylo; May 17th, 2017 at 07:59 AM. 

Tags 
complex 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Basis for Complex Vector Space over the Complex and then the Real Fields  whh2  Linear Algebra  1  April 30th, 2015 04:00 AM 
Can complex numbers (and properties of complex numbers) be..  jonas  Complex Analysis  2  October 13th, 2014 03:03 PM 
Complex complex numbers  Tutu  Algebra  11  June 26th, 2012 01:36 PM 
complex no.  panky  Algebra  1  August 2nd, 2011 10:52 PM 
Complex Solutions to A "Complex" quadratic  Ajihood  Complex Analysis  1  March 16th, 2009 09:02 AM 