The value of 2-i 2-i1-i 1-i is- where i2-1

We have, 2+i amp;2 i 1+i amp;1 i = (2+i)(1 i) (2 i)(1+i) =2 2i+i i^2 2 2i+i+i^2 = 2i ...

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Multiplication with Vectors

The value of ...

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The value of $∣ ∣ ∣ \begin{matrix} 2 + i & 2 - i 1 + i & 1 - i \end{matrix} ∣ ∣ ∣$ is, where $(i^{2} = - 1)$

purely imaginary

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a real quantity

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1 + i

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1 - i

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Find the value of $∣ ∣ ∣ \begin{matrix} 2 + i & 2 - i 1 + i & 1 - i \end{matrix} ∣ ∣ ∣$ if $i^{2} = - 1$ .

∣ ∣ ∣ \begin{matrix} 2 + i & 2 - i 1 + i & 1 - i \end{matrix} ∣ ∣ ∣

f (x)

I_{1} = \int_{0}^{\infty} f (x) d x,

I_{2} = \int_{0}^{\infty} f (3 x - \frac{1}{2 x}) d x,

\frac{I_{1}}{I_{2}}

I_{1} & I_{2}

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

z

L - C - R

$(A) .$ If $ω L_{1} - \frac{1}{C_{1} ω} = R_{1}$ , and $ω L_{2} - \frac{1}{ω C_{2}} = R_{2}$	$(1) .$ $I_{1}$ and $I_{2}$ are in same Phase.
$(B) .$ If $ω L_{1} - \frac{1}{C_{1} ω} = R_{1}$ , and $\frac{1}{ω C_{2}} - ω L_{2} = R_{2}$	$(2) .$ $I = I_{1} + I_{2}$ where , $I_{1}, I_{2}, I_{3}$ are RMS value of currents.
$(C) .$ If capacitor $C_{1}$ and inductor $L_{2}$ are removed from circuit and $ω L_{1} = R_{1}; \frac{1}{ω C_{2}} = R_{2}$	$(3) .$ Magnitude of phase difference between $I_{1}$ and $I_{2}$ is $\frac{π}{2}$
$(D) .$ If capacitors $C_{1}$ and $C_{2}$ are both removed from the circuit and $ω L_{1} = R_{1}; ω L_{2} = R_{2}$	$(4) .$ $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{2})}_{2}^{2}}$ where, $I_{1}, I_{2}, I_{3}$ are RMS value of currents.
$(5) .$ $I_{1}$ is lags and $I_{2}$ is leads from source voltage.

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