# Complex number equation graph problem

• MHB
• jaychay
In summary, you are looking for the equation for Zo and the coordinate (a,b) in a coordinate system. You are given that Zo is equal to (1,1) and that (a,b) is located in the shaded area in the opening post.
jaychay

Given (a,b) is the coordinate just like (x,y). Find equation Zo and coordinate (a,b) ?Please help me

Last edited:
Is the answer $z_o = (1,1)$ or simply $1+i$ and $(a, b)=(2, 1.8)$ ? Not sure about b though

DaalChawal said:
Is the answer $z_o = (1,1)$ or simply $1+i$ and $(a, b)=(2, 1.8)$ ? Not sure about b though
Can you tell me where does (a,b) = ( 2,1.8 ) come from ?

first of all $(a, b)$ are not co-ordinates... a represents the radius of circle centered at $z_o$ .
For b see the graph... from the circle we can say z must be outside the circle then it's real part should also be outside it(see the minimum value of Re(z) so that it is outside the circle)

I am really struggle with question 2 on how to find (a,b) I am not sure that b = 1.8 is correct or not

jaychay said:
I am really struggle with question 2 on how to find (a,b) I am not sure that b = 1.8 is correct or not
We are given the inequality $\operatorname{Re}z\ge b$ and in the graph we can see that the shaded area is to the right of the imaginary axis.
Therefore $b=0$.

Klaas van Aarsen said:
We are given the inequality $\operatorname{Re}z\ge b$ and in the graph we can see that the shaded area is to the right of the imaginary axis.
Therefore $b=0$.
So ( a,b ) is equal to ( 2,0 ) right ?

jaychay said:
So ( a,b ) is equal to ( 2,0 ) right ?
Yes.

Klaas van Aarsen said:
We are given the inequality $\operatorname{Re}z\ge b$ and in the graph we can see that the shaded area is to the right of the imaginary axis.
Therefore $b=0$.
Sir you are saying that Re(z) = 0 that means z =0 + i y means z will lie on imazinary axis but from the graph z lies in 1st and 4th quadrants except inside the circle. How is this possible?

Klaas van Aarsen said:
Yes.
On question one Zo is equal to (1,1) right ?

DaalChawal said:
Sir you are saying that Re(z) = 0 that means z =0 + i y means z will lie on imazinary axis but from the graph z lies in 1st and 4th quadrants except inside the circle. How is this possible?
We don't have $\operatorname{Re}(z)=0$ for the shaded area. Instead we have $\operatorname{Re}(z)\ge 0$.

[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-6.756756756756757,"xmax":10,"ymax":6.756756756756757}},"randomSeed":"fb6b112a792f74beca093fcec90d2a8f","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"x\\ge0"},{"type":"expression","id":"3","color":"#388c46"}]}}[/DESMOS]

The combination of $|z-(1+i)|>2$ and $\operatorname{Re} z\ge 0$ shows up as:

[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-6.756756756756757,"xmax":10,"ymax":6.756756756756757}},"randomSeed":"5dafa6730d6be2f3a851403cb787a49a","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"\\left(x-1\\right)^{2}+\\left(y-1\\right)^{2}>2^{2}\\left\\{x\\ge0\\right\\}"},{"type":"expression","id":"2","color":"#2d70b3"},{"type":"expression","id":"3","color":"#388c46"}]}}[/DESMOS]

As we can see, this matches the shaded area in the opening post.

jaychay said:
On question one Zo is equal to (1,1) right ?
I'd make it $z_0=1+i$, since $z_0$ is an imaginary number.
To be fair, that is equivalent to $(1,1)$, which is the cartesian representation of the same imaginary number.

## 1. What is a complex number equation graph problem?

A complex number equation graph problem is a mathematical problem that involves plotting the solutions of a complex number equation on a graph. Complex numbers are numbers that consist of a real part and an imaginary part, and they are represented as a + bi, where a and b are real numbers and i is the imaginary unit.

## 2. How do you graph a complex number equation?

To graph a complex number equation, you can plot the real part of the complex number on the x-axis and the imaginary part on the y-axis. The resulting point on the graph represents the solution of the equation. You can also use the distance formula to find the distance between the origin and the point on the graph, which represents the magnitude of the complex number.

## 3. What is the purpose of graphing complex number equations?

Graphing complex number equations can help visualize the solutions of the equation and understand the relationship between the real and imaginary parts. It can also help in solving equations and identifying patterns or trends.

## 4. How do you solve a complex number equation graph problem?

To solve a complex number equation graph problem, you can use algebraic methods such as factoring, completing the square, or the quadratic formula. You can also use graphical methods, such as finding the intersection points of the graph with the x-axis or using the distance formula to find the magnitude of the complex number.

## 5. What are some real-life applications of complex number equation graph problems?

Complex number equation graph problems have many real-life applications, including in engineering, physics, and economics. They can be used to model and analyze systems with both real and imaginary components, such as electrical circuits, mechanical systems, and financial markets.

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