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Post a LessonAnswered on 14 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/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 CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/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 CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/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 CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/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 CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/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.
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