$4 a b - 2 (a^{2} - b^{2}) i = 0 + 0 i$ Find a,b.

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Solution

Given,

$4 a b - 2 (a^{2} - b^{2}) i = 0$

$4 a b + (- 2 a^{2} + 2 b^{2}) i = 0 + 0 i$

$[\begin{matrix} 4 a b = 0 - 2 a^{2} + 2 b^{2} = 0 \end{matrix}]$

solving the above 2 equations, we get,

$b = 0, a = 0$

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Similar questions

\sqrt{4 a b - 2 (a^{2} - b^{2}) i}

4 a b - 2 (a^{2} - b^{2}) i

△ A B C

2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c,

∠ A

∠ B

(a + b)^{2} - (a - b)^{2}

4 a b

- 4 a b

2 a^{2} + 2 b^{2}

2 a^{2} - 2 b^{2}

A B C, 2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c,

cos B

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