If alpha, beta are non-zero integers and z=(a | vedclass

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(B) Given $z = (\alpha + i\beta)(2 + 7i) = (2\alpha - 7\beta) + i(7\alpha + 2\beta)$.Since $z$ is purely imaginary,the real part must be zero:$2\alpha

Vedclass · Accepted Answer

$2809$

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(B) Given $z = (\alpha + i\beta)(2 + 7i) = (2\alpha - 7\beta) + i(7\alpha + 2\beta)$.Since $z$ is purely imaginary,the real part must be zero:$2\alpha - 7\beta = 0 \Rightarrow 2\alpha = 7\beta$.Since $\alpha, \beta$ are integers,let $\alpha = 7k$ and $\beta = 2k$ for some non-zero integer $k$.Then $|z|^2 = (	ext{Re}(z))^2 + (	ext{Im}(z))^2 = 0^2 + (7\alpha + 2\beta)^2$.Substituting $\alpha = 7k$ and $\beta = 2k$:$|z|^2 = (7(7k) + 2(2k))^2 = (49k + 4k)^2 = (53k)^2 = 2809k^2$.For the minimum non-zero value,we take $k = 1$ (or $k = -1$):$|z|^2 = 2809(1)^2 = 2809$.

If $\alpha, \beta$ are non-zero integers and $z=(\alpha+i \beta)(2+7 i)$ is a purely imaginary number,then the minimum value of $|z|^2$ is

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