Explain the fallacy in the following- - 1 - i - i - - - 1 - - 1 - - - 1 - 1 - - 1 - 1

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Standard XII

Mathematics

Complex Numbers

Explain the f...

Question

Explain the fallacy in the following:
$- 1 = i, i = \sqrt{- 1} \sqrt{- 1} = \sqrt{(- 1) (- 1)} = \sqrt{1} = 1$

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Solution

$\sqrt{- 1} \sqrt{- 1} \neq \sqrt{(- 1) (- 1)}$
The property $s q r t a \sqrt{b} = \sqrt{a b}$ is valid only if at least one of a and b is non-negative.
If a and b are both negative then $\sqrt{a} \sqrt{b} = - \sqrt{| a | | b |}$ is used.

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Similar questions

Q. Simplify the following :

\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}

Q. Express the following complex numbers in the standard form a + i b:
(i)

(1 + i) (1 + 2 i)

(ii)

\frac{3 + 2 i}{- 2 + i}

(iii)

\frac{1}{(2 + i)^{2}}

(iv)

\frac{1 - i}{1 + i}

(v)

\frac{(2 + i)^{3}}{2 + 3 i}

(vi)

\frac{(1 + i) (1 + \sqrt{3} i)}{1 - i}

(vii)

\frac{2 + 3 i}{4 + 5 i}

(viii)

\frac{(1 - i)^{3}}{1 - i^{3}}

(ix)

(1 + 2 i)^{- 3}

(x)

\frac{3 - 4 i}{(4 - 2 i) (1 + i)}

(xi)

(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{1 - 4 i}{5 + i})

(xii)

\frac{5 + \sqrt{2} i}{1 - 2 \sqrt{i}}

Convert the following in the polar form:

(i) , (ii)

Convert the following in the polar form:

$(i) \frac{1 + 7 i}{(2 - i)^{2}}$

$(i i) \frac{1 + 3 i}{1 - 2 i}$

Express the following complex numbers in the standard form a + i b :

$(i) (1 + i) (1 + 2 i) (i i) \frac{3 + 2 i}{- 2 + i} (i i i) \frac{1}{(2 + i)^{2}} (i v) \frac{1 - i}{1 + i} (v) \frac{(2 + i)^{3}}{2 + 3 i} (v i) \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} (v i i) \frac{2 + 3 i}{4 + 5 i} (v i i i) \frac{(1 - i)^{3}}{1 - i^{3}} (i x) (1 + 2 i)^{- 3} (x) \frac{3 - 4 i}{(4 - 2 i) (1 + i)} (x i) (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) (x i i) \frac{5 + \sqrt{2 i}}{1 - \sqrt{2 i}}$

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Explain the fallacy in the following: −1=i,i=√−1√−1=√(−1)(−1)=√1=1

√−1√−1≠√(−1)(−1)The property sqrta√b=√abis valid only if at least one of a and b is non-negative.If a and b are both negative then √a√b=−√|a||b| is used.

Explain the fallacy in the following:
$- 1 = i, i = \sqrt{- 1} \sqrt{- 1} = \sqrt{(- 1) (- 1)} = \sqrt{1} = 1$

$\sqrt{- 1} \sqrt{- 1} \neq \sqrt{(- 1) (- 1)}$
The property $s q r t a \sqrt{b} = \sqrt{a b}$ is valid only if at least one of a and b is non-negative.
If a and b are both negative then $\sqrt{a} \sqrt{b} = - \sqrt{| a | | b |}$ is used.