For real numbers a and b,if 4a + i(3a - b) = | vedclass

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(C) Given the equation: $4a + i(3a - b) = b - 6i$ Equating the real and imaginary parts on both sides: Real part: $4a = b$ Imaginary part: $3a - b = -

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$\sqrt{2}$

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(C) Given the equation: $4a + i(3a - b) = b - 6i$ Equating the real and imaginary parts on both sides: Real part: $4a = b$ Imaginary part: $3a - b = -6$ Substitute $b = 4a$ into the second equation: $3a - 4a = -6$ $\Rightarrow -a = -6$ $\Rightarrow a = 6$ Then,$b = 4(6) = 24$ Now,substitute $a$ and $b$ into $z$: $z = 6 + \frac{24}{4}i = 6 + 6i$ The modulus $|z|$ is: $|z| = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$ Finally,calculate $\frac{|z|}{a}$: $\frac{|z|}{a} = \frac{6\sqrt{2}}{6} = \sqrt{2}$

For real numbers $a$ and $b$,if $4a + i(3a - b) = b - 6i$ and $z = a + \frac{b}{4}i$,then $\frac{|z|}{a} = $

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