My Math Forum  

Go Back   My Math Forum > College Math Forum > Complex Analysis

Complex Analysis Complex Analysis Math Forum


Thanks Tree8Thanks
Reply
 
LinkBack Thread Tools Display Modes
May 1st, 2017, 06:29 AM   #1
Senior Member
 
Joined: Nov 2015
From: hyderabad

Posts: 148
Thanks: 1

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.
Lalitha183 is offline  
 
May 1st, 2017, 08:13 AM   #2
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,549
Thanks: 1260

$z = a+bi$

$|z| = 5 \implies \sqrt{a^2+b^2} = 5 \implies a^2+b^2 = 25$

$z+1 = a+bi + 1 = (a+1) + bi$

$z+i = a+bi + i = a + (b+1)i$

$|z+1| = |z+i| \implies \sqrt{(a+1)^2 + b^2} = \sqrt{a^2 + (b+1)^2}$

$(a+1)^2 + b^2 = a^2 + (b+1)^2$

$a^2+2a+1+b^2 = a^2+b^2+2b+1$

$2a=2b \implies a=b$

$a^2+b^2 = 25 \implies 2a^2 = 25 \implies a = b = \pm \dfrac{5\sqrt{2}}{2}$

so, how many solutions for $z$ ?
skeeter is offline  
May 1st, 2017, 12:15 PM   #3
Senior Member
 
Joined: Aug 2012

Posts: 1,376
Thanks: 327

$\lvert z \rvert = 5$ 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.
Maschke is offline  
May 1st, 2017, 01:04 PM   #4
Senior Member
 
Joined: Aug 2012

Posts: 1,376
Thanks: 327

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.
Maschke is offline  
May 1st, 2017, 05:13 PM   #5
Senior Member
 
Joined: Aug 2012

Posts: 1,376
Thanks: 327

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.
Maschke is offline  
May 1st, 2017, 11:02 PM   #6
Senior Member
 
Joined: Aug 2012

Posts: 1,376
Thanks: 327

New improved artwork. No more I promise

Thanks from agentredlum
Maschke is offline  
May 2nd, 2017, 01:59 AM   #7
Math Team
 
agentredlum's Avatar
 
Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,261
Thanks: 198

I support the use of pictures in mathematics.

agentredlum is offline  
May 10th, 2017, 04:13 AM   #8
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,488
Thanks: 630

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.
Country Boy is online now  
May 10th, 2017, 04:30 AM   #9
Senior Member
 
Joined: Nov 2015
From: hyderabad

Posts: 148
Thanks: 1

Quote:
Originally Posted by Country Boy View Post
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
Lalitha183 is offline  
May 17th, 2017, 07:42 AM   #10
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,056
Thanks: 85

(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.
zylo is offline  
Reply

  My Math Forum > College Math Forum > Complex Analysis

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





Copyright © 2017 My Math Forum. All rights reserved.