Let
\(S = \left\{z∈C : z^2+\overline{z} = 0 \right\}\). Then \(∑_{z∈S}(Re(z)+Im(z))\)
is equal to____.
The correct answer is 0
\(∵ z^2+\overline{z} = 0\) Let \(z = x+iy\)
\(∴ x^2+y^2+2ixy+x-iy = 0\)
\((x^2-y^2+x)+i(2xy-y) = 0\)
\(∴ x^2+y^2 = 0\) and \((2x-1)y = 0\)
If \(x = +\frac{1}{2}\) then \(y = ±\frac{\sqrt3}{2}\)
And if y = 0 then x = 0, –1
\(∴\) \(z = \{ 0 + 0i, -1 + 0i, \frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i \}\)
\(∴ ∑(R_e(z)+m(z)) = 0\)
If \( \text{Re} \left( \frac{2z + i}{z + i} \right) + \text{Re} \left( \frac{2z - i}{z - i} \right) = 2 \) is a circle of radius \( r \) and centre \( (a, b) \), then \( \frac{15ab}{r^2} \) is equal to:
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $
A Complex Number is written in the form
a + ib
where,
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.