In grade 10 math , complex conjugates are a pair of complex numbers, both having the same real part, but with imaginary parts of opposite signs. For example, 3 + 4i and 3 - 4i are complex conjugates.
The conjugate of the complex number z
notation avoids confusion with the notation for the conjugate transpose of a matrix, which can be thought of as a generalization of complex conjugation. If a complex number is represented as a 2×2 matrix, the notations are identical.We also can use scientific notation converter to find the number.
Let's an example problem from numeric and algebraic operations
Question:-
How do you use complex conjugates to find (3+7i)/(2-i) ?
Answer:-
In the Given problem the denominator is (2-i)
So ,it's conjugate is (2+i)
Multiplying both numerator and denominator with (2+i)
= (3+7i)(2+i)/(2-i)(2+i)
= (5+17i)/(4-i2)
We know that i2 = -1
= (5+17i)/4+1
= (5+17i)/5 Answer
The conjugate of the complex number z
notation avoids confusion with the notation for the conjugate transpose of a matrix, which can be thought of as a generalization of complex conjugation. If a complex number is represented as a 2×2 matrix, the notations are identical.We also can use scientific notation converter to find the number.Let's an example problem from numeric and algebraic operations
Question:-
How do you use complex conjugates to find (3+7i)/(2-i) ?
Answer:-
In the Given problem the denominator is (2-i)
So ,it's conjugate is (2+i)
Multiplying both numerator and denominator with (2+i)
= (3+7i)(2+i)/(2-i)(2+i)
= (5+17i)/(4-i2)
We know that i2 = -1
= (5+17i)/4+1
= (5+17i)/5 Answer
![f(x,y) = xe^{x^2y}\\\frac{\partial f}{\partial x}=? And \frac{\partial f}{\partial y}=?\\\frac{\partial f}{\partial x}_{(1,2)}=? And \frac{\partial f}{\partial y}_{(1,ln 2)}=?\\ f(x,y) = xe^{x^2y}<br />\\Consider \frac{\partial e^{x^2y}}{\partial x}\\\frac{\partial e^u}{\partial x} and u=x^2y\\\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\ as u=x^2y\\\frac{\partial e^u}{\partial u}=e^u=e^{x^2y}----------(a)\\\frac{\partial u}{\partial x}=2xy-------(b)\\ By Chain Rule\\\frac{\partial e^{x^2y}}{\partial x}=\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\=e^{x^2y} . 2xy where a = e^{x^2y} and b=2xy\\\frac{\partial f}{\partial x}=x \frac{\partial e^{x^2y}}{\partial x}+e^{x^2y}\frac{\partial x}{\partial x}\\=xe^{x^2y}(2xy)+e^{x^2y}(1)\\=e^{x^2y}[2x^2y+1]\\\frac{\partial f}{\partial x}=xe^{x^2y}(x^2)\\=x^3e^{x^2y}\\Therefore\frac{\partial f}{\partial x}_{(1,ln2)}=e^{1ln2}(i^2.ln2)=e^{ln2}=2ln2\\\frac{\partial f}{\partial y}_{(1,ln2)}=1^3 e^{1^2ln(2)}=e^{ln(2)}=2<br />](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDA9G1EQoYK18sd1GibXDdHa8cpFAP50o9rwzNzbmdv5Xa2-Uz0wNPFBXtUc9AUkcbzkYKYksefcitwhs8g46nGNw6BAgNsijf7mJuiJwxi4syAbe7q8miLe71y4VTOX0_OET_ktTX4cM/s1600-h/finaldydx.JPG)
