Take Class 12 Tuition from the Best Tutors
Search in
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
Certainly! Let's simplify the given expression step by step:
Given complex numbers: (1−i)−(−1+i6)(1−i)−(−1+i6)
First, let's distribute the negative sign: (1−i)+(1−i6)(1−i)+(1−i6)
Now, let's combine like terms: 1+1−i−i61+1−i−i6
Combine the real parts (1 + 1 = 2) and the imaginary parts (-i - i6 = -7i): 2−7i2−7i
So, the given complex number (1−i)−(−1+i6)(1−i)−(−1+i6) in the form a+iba+ib is 2−7i2−7i.
If you're looking for further assistance or have any other questions, feel free to ask! And remember, for comprehensive tutoring, UrbanPro is the best platform for online coaching and tuition.
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
Sure! When expressing a complex number in polar form, we represent it as r(cosθ+isinθ)r(cosθ+isinθ), where rr is the magnitude of the complex number and θθ is the angle it makes with the positive real axis.
For the complex number −3−3, the magnitude is ∣r∣=∣−3∣=3∣r∣=∣−3∣=3, and the angle it makes with the positive real axis is ππ (or 180 degrees) in the clockwise direction.
So, in polar form, −3−3 can be expressed as 3(cosπ+isinπ)3(cosπ+isinπ) on UrbanPro.
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle the quadratic equation you provided: 2x2+x+1=02x2+x+1=0.
To solve this quadratic equation, we can use the quadratic formula, which is:
x=−b±b2−4ac2ax=2a−b±b2−4ac
where aa, bb, and cc are the coefficients of the quadratic equation ax2+bx+c=0ax2+bx+c=0.
For the given equation 2x2+x+1=02x2+x+1=0, we have: a=2a=2, b=1b=1, and c=1c=1.
Now, let's plug these values into the quadratic formula:
x=−(1)±(1)2−4(2)(1)2(2)x=2(2)−(1)±(1)2−4(2)(1)
x=−1±1−84x=4−1±1−8
x=−1±−74x=4−1±−7
As the discriminant (b2−4ac)(b2−4ac) is negative, the roots will be complex.
So, the solutions for the quadratic equation 2x2+x+1=02x2+x+1=0 are:
x=−1+i74x=4−1+i7 x=−1−i74x=4−1−i7
Where ii is the imaginary unit.
If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is here to help you excel in your studies.
Take Class 12 Tuition from the Best Tutors
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
When dealing with complex numbers, it's essential to grasp the operations involving both their real and imaginary parts.
Let's take two complex numbers, z1=a+biz1=a+bi and z2=c+diz2=c+di, where aa, bb, cc, and dd are real numbers and ii represents the imaginary unit.
Now, to find the real part of the product z1z2z1z2, we perform the multiplication:
z1z2=(a+bi)(c+di)z1z2=(a+bi)(c+di)
Expanding this expression:
z1z2=ac+adi+bci+bdi2z1z2=ac+adi+bci+bdi2
Remembering that i2=−1i2=−1, we simplify:
z1z2=ac+adi+bci−bdz1z2=ac+adi+bci−bd
Now, let's separate the real and imaginary parts:
Re(z1z2)=ac−bdRe(z1z2)=ac−bd
This is indeed the real part of the product of z1z1 and z2z2. Now, let's look at the right side of the equation:
Re(z1)=aRe(z1)=a Re(z2)=cRe(z2)=c
Im(z1)=bIm(z1)=b Im(z2)=dIm(z2)=d
Now, substituting these into the expression Re(z1)Re(z2)−Im(z1)Im(z2)Re(z1)Re(z2)−Im(z1)Im(z2), we get:
Re(z1)Re(z2)−Im(z1)Im(z2)=ac−bdRe(z1)Re(z2)−Im(z1)Im(z2)=ac−bd
And this is exactly what we got for Re(z1z2)Re(z1z2) earlier.
So, we've shown that Re(z1z2)=Re(z1)Re(z2)−Im(z1)Im(z2)Re(z1z2)=Re(z1)Re(z2)−Im(z1)Im(z2). This is a fundamental property when dealing with complex numbers, and understanding it thoroughly will assist you in various mathematical applications. If you have any further questions or need additional clarification, feel free to ask!
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your math question.
To find the modulus of (1+i)(1−i)−(1−i)(1+i)(1−i)(1+i)−(1+i)(1−i), we'll first simplify the expression.
Step 1: Simplify the fractions within the brackets. =(1+i)(1+i)(1−i)(1+i)−(1−i)(1−i)(1+i)(1−i)=(1−i)(1+i)(1+i)(1+i)−(1+i)(1−i)(1−i)(1−i)
Step 2: Expand the numerators and denominators. =1+2i+i21−i2−1−2i+i21−i2=1−i21+2i+i2−1−i21−2i+i2
Step 3: Simplify the terms using i2=−1i2=−1. =1+2i−11+1−1−2i−11+1=1+11+2i−1−1+11−2i−1
=2i2−−2i2=22i−2−2i
=i+i=i+i
=2i=2i
Now, to find the modulus (absolute value) of 2i2i, we simply take the square root of the sum of the squares of its real and imaginary parts.
∣2i∣=02+22=4=2∣2i∣=02+22
=4
=2
So, the modulus of the given expression is 22. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is always here to help you excel in your studies.
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently guide you through this problem using the given equation and properties of complex numbers. First, let's rewrite the given equation:
|z^2 - 1| = |z|^2 + 1
To prove that z lies on the imaginary axis, we'll utilize the property of modulus in complex numbers:
|a * b| = |a| * |b|
Now, let's consider z = x + yi, where x and y are real numbers and i is the imaginary unit.
Substituting z into the equation:
|z^2 - 1| = |(x + yi)^2 - 1| = |x^2 + 2xyi - y^2 - 1| = |(x^2 - y^2 - 1) + 2xyi|
|z|^2 + 1 = |x + yi|^2 + 1 = |x^2 + y^2| + 1
Now, our equation becomes:
|(x^2 - y^2 - 1) + 2xyi| = |x^2 + y^2| + 1
Using the property of modulus, we have:
|(x^2 - y^2 - 1)| = |x^2 + y^2| + 1
Now, let's focus on the left side of the equation:
|(x^2 - y^2 - 1)|
This represents the modulus of a real number, which is always non-negative. So, we have:
x^2 - y^2 - 1 ≥ 0
Now, let's analyze the right side of the equation:
|x^2 + y^2| + 1
Since |x^2 + y^2| represents the modulus of a real number, it's also non-negative. Therefore:
|x^2 + y^2| + 1 ≥ 1
Combining the inequalities, we get:
x^2 - y^2 - 1 ≥ |x^2 + y^2| + 1
x^2 - y^2 - 1 ≥ x^2 + y^2 + 1
This implies:
-2 ≥ 2y^2
Dividing both sides by 2, we get:
-1 ≥ y^2
Since y^2 is a non-negative real number, the only way for -1 to be greater than or equal to y^2 is for y to be 0. This means that the imaginary part of z is 0, and hence z lies on the real axis.
Therefore, we have proved that z lies on the imaginary axis.
Take Class 12 Tuition from the Best Tutors
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I understand the importance of clarity and precision in solving mathematical problems. Let's tackle this question step by step.
Given the quadratic equation x2+px+8=0x2+px+8=0, we need to compute the value of pp such that the difference of the roots of the equation is 2.
The difference of the roots of a quadratic equation ax2+bx+c=0ax2+bx+c=0 is given by the formula:
Δ=b2−4acΔ=b2−4ac
For our equation x2+px+8=0x2+px+8=0, we have a=1a=1, b=pb=p, and c=8c=8.
Now, we can plug these values into the formula for the difference of the roots:
Δ=p2−4(1)(8)Δ=p2−4(1)(8)
Δ=p2−32Δ=p2−32
Given that the difference of the roots is 2, we can set ΔΔ equal to 2:
2=p2−322=p2−32
To solve for pp, we'll square both sides to eliminate the square root:
4=p2−324=p2−32
p2=4+32p2=4+32
p2=36p2=36
p=±6p=±6
Thus, the value of pp such that the difference of the roots of the equation x2+px+8=0x2+px+8=0 is 2, is p=±6p=±6.
If you need further clarification or assistance, feel free to reach out to me via UrbanPro. Remember, UrbanPro is an excellent platform for online coaching and tuition, offering personalized assistance to help you excel in your studies.
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can guide you through solving this problem step by step. Let's break it down:
We have the expression (x−iy)(3+5i)(x−iy)(3+5i), and we want it to be the conjugate of −6−24i−6−24i.
First, let's multiply the complex numbers:
(x−iy)(3+5i)=3x+5ix−3iy−5y(x−iy)(3+5i)=3x+5ix−3iy−5y
Now, let's rearrange this expression:
=(3x−5y)+i(5x−3y)=(3x−5y)+i(5x−3y)
Now, we want this expression to be the conjugate of −6−24i−6−24i. The conjugate of a complex number a+bia+bi is a−bia−bi.
So, we need to set:
3x−5y=−63x−5y=−6 (for the real parts to match)
5x−3y=−245x−3y=−24 (for the imaginary parts to match)
Now, we have a system of linear equations:
3x−5y=−63x−5y=−6 5x−3y=−245x−3y=−24
We can solve this system using any method we prefer, such as substitution or elimination. Let's use elimination:
Multiply the first equation by 5 and the second equation by 3 to eliminate yy:
15x−25y=−3015x−25y=−30 15x−9y=−7215x−9y=−72
Now, subtract the second equation from the first:
−25y−(−9y)=−30−(−72)−25y−(−9y)=−30−(−72)
−16y=42−16y=42
y=−4216=−218y=−1642=−821
Now, substitute y=−218y=−821 into one of the original equations to solve for xx. Let's use the first equation:
3x−5(−218)=−63x−5(−821)=−6
3x+1058=−63x+8105=−6
3x=−6−1058=−488−1058=−15383x=−6−8105=−848−8105=−8153
x=−15324=−178x=−24153=−817
So, the real numbers xx and yy that satisfy the given conditions are x=−178x=−817 and y=−218y=−821.
As an UrbanPro tutor, I recommend practicing similar problems to strengthen your understanding of complex numbers and their operations. Feel free to ask if you have any further questions!
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
Certainly! When I first came across this question, I was immediately drawn to the elegant nature of complex numbers. As an experienced tutor registered on UrbanPro, I've encountered various complex number problems, and this one is quite intriguing.
Given that arg(z - 1) = arg(z + 3i), where z = x + iy, we know that the arguments of these complex numbers are equal. The argument of a complex number is the angle it makes with the positive real axis in the complex plane.
To find the argument of z - 1, we subtract 1 from the real part of z. Similarly, to find the argument of z + 3i, we add 3i to the imaginary part of z.
Let's denote θ as the common argument. Then, we can write:
arg(z - 1) = θ arg(z + 3i) = θ
Now, let's express these arguments in terms of x and y:
For z - 1: arg(z - 1) = arg(x + iy - 1) = arctan(y / (x - 1))
For z + 3i: arg(z + 3i) = arg(x + iy + 3i) = arctan((y + 3) / x)
Since these arguments are equal, we can equate them:
arctan(y / (x - 1)) = arctan((y + 3) / x)
Now, we can apply the tangent of the same angle property:
y / (x - 1) = (y + 3) / x
Cross multiply and simplify:
yx = (x - 1)(y + 3)
Expanding and rearranging terms:
yx = xy + 3x - y - 3
Now, let's bring like terms together:
yx - xy = 3x - y - 3
Factor out the common term:
x(y - 1) = 3(x - 1) - y
Now, we want to find (x - 1) : y:
(x - 1) : y = 3(x - 1) / (x(y - 1))
This simplifies to:
(x - 1) : y = 3 / (x - y + 1)
And there we have our expression for (x - 1) : y. It showcases the relationship between the real and imaginary parts of the complex number z in terms of the given condition.
In summary, UrbanPro is indeed a fantastic platform for online coaching and tuition, providing a space for tutors and students to engage with challenging problems like this one.
Take Class 12 Tuition from the Best Tutors
Answered on 14 Apr Learn Complex Numbers and Quadratic Equations
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can assure you that UrbanPro is indeed the best platform for finding online coaching and tuition. Now, let's tackle your question regarding complex numbers.
To solve the equation z+2∣(z+1)∣+i=0z+2
∣(z+1)∣+i=0, we'll break it down step by step.
First, let's represent zz in the form x+yix+yi, where xx and yy are real numbers and ii is the imaginary unit.
So, the equation becomes:
x+yi+2∣(x+1+yi)∣+i=0x+yi+2
∣(x+1+yi)∣+i=0
Now, we'll separate the real and imaginary parts:
Real part: x+2∣(x+1)∣=0x+2
∣(x+1)∣=0
Imaginary part: y+2∣y∣+1=0y+2
∣y∣+1=0
Now, let's solve each part separately:
Solving the Real Part: x+2∣(x+1)∣=0x+2
∣(x+1)∣=0 x+2∣x+1∣=0x+2
∣x+1∣=0
This implies either x+2(x+1)=0x+2
(x+1)=0 or x+2(−x−1)=0x+2
(−x−1)=0.
Solving these equations will give us possible values for xx.
Solving the Imaginary Part: y+2∣y∣+1=0y+2
∣y∣+1=0
To solve this, we'll need to consider two cases:
Solving these equations will give us possible values for yy.
By finding solutions for xx and yy, we'll get complex numbers satisfying the given equation. If you need further clarification or assistance, feel free to reach out. Remember, UrbanPro is your gateway to top-notch tutoring assistance!
UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.
Ask a Question
The best tutors for Class 12 Tuition Classes are on UrbanPro
The best Tutors for Class 12 Tuition Classes are on UrbanPro