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

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Aicharya · Accepted Answer

$\because 4 a+i(3 a-b)=b-6 i$ Comparing both sides, we get : $\begin{aligned} & 4 a=b \ & 3 a-b=-6 \Rightarrow 3 a-4 a=-6 \Rightarrow, a=6 \ & 	herefore \quad b=24 \ & Z=a+\frac{b}{4} i \Rightarrow z=6+\frac{24}{4} i=6+6 i \ & |z|=\sqrt{36+36}=6 \sqrt{2} \ & \frac{|z|}{a}=\frac{6 \sqrt{2}}{6}=\sqrt{2} \end{aligned}$

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

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