
How do you find $\left| 6-3i \right|$ ?
Answer
533.7k+ views
Hint: To find the value of $\left| 6-3i \right|$, we are going to use the property of a complex number because the term written inside the modulus is a complex number. We know that $\left| a+ib \right|$ is the magnitude of the complex number $a+ib$ and magnitude is calculated as follows: $\sqrt{{{a}^{2}}+{{b}^{2}}}$. In this way, we can find the value of $\left| 6-3i \right|$.
Complete step by step solution:
In the above problem, we have given the following:
$\left| 6-3i \right|$
And we are asked to calculate the value of this modulus. The term written inside the modulus is a complex number because it is in the form of $a+ib$. We know that applying modulus on any complex number is signifying that we have to find the magnitude of the complex number.
Magnitude of complex number $a+ib$ is equal to:
$\sqrt{{{a}^{2}}+{{b}^{2}}}$
So, applying the above property of finding the magnitude of the complex number $6-3i$ we get,
$\begin{align}
& \left| 6-3i \right|=\sqrt{{{6}^{2}}+{{\left( -3 \right)}^{2}}} \\
& \Rightarrow \left| 6-3i \right|=\sqrt{36+9} \\
& \Rightarrow \left| 6-3i \right|=\sqrt{45} \\
\end{align}$
To simplify further we are going to factorize 45 we get,
$\begin{align}
& \left| 6-3i \right|=\sqrt{9\times 5} \\
& \Rightarrow \left| 6-3i \right|=3\sqrt{5} \\
\end{align}$
Hence, we have found the value of $\left| 6-3i \right|=3\sqrt{5}$
Note: The mistake that could be possible in the above problem is that while finding the magnitude of the complex number you might forgot to put the square root sign and then it will look like as follows:
$\left| 6-3i \right|=\left( {{6}^{2}}+{{\left( -3 \right)}^{2}} \right)$
This is the incorrect formula for finding the magnitude of the complex. The above formula will be true if we put square sign on the L.H.S of the above equation which we are shown below:
${{\left| 6-3i \right|}^{2}}=\left( {{6}^{2}}+{{\left( -3 \right)}^{2}} \right)$
But to find the magnitude of the complex number always put square roots and then do the square of the real and imaginary part and then add them.
Complete step by step solution:
In the above problem, we have given the following:
$\left| 6-3i \right|$
And we are asked to calculate the value of this modulus. The term written inside the modulus is a complex number because it is in the form of $a+ib$. We know that applying modulus on any complex number is signifying that we have to find the magnitude of the complex number.
Magnitude of complex number $a+ib$ is equal to:
$\sqrt{{{a}^{2}}+{{b}^{2}}}$
So, applying the above property of finding the magnitude of the complex number $6-3i$ we get,
$\begin{align}
& \left| 6-3i \right|=\sqrt{{{6}^{2}}+{{\left( -3 \right)}^{2}}} \\
& \Rightarrow \left| 6-3i \right|=\sqrt{36+9} \\
& \Rightarrow \left| 6-3i \right|=\sqrt{45} \\
\end{align}$
To simplify further we are going to factorize 45 we get,
$\begin{align}
& \left| 6-3i \right|=\sqrt{9\times 5} \\
& \Rightarrow \left| 6-3i \right|=3\sqrt{5} \\
\end{align}$
Hence, we have found the value of $\left| 6-3i \right|=3\sqrt{5}$
Note: The mistake that could be possible in the above problem is that while finding the magnitude of the complex number you might forgot to put the square root sign and then it will look like as follows:
$\left| 6-3i \right|=\left( {{6}^{2}}+{{\left( -3 \right)}^{2}} \right)$
This is the incorrect formula for finding the magnitude of the complex. The above formula will be true if we put square sign on the L.H.S of the above equation which we are shown below:
${{\left| 6-3i \right|}^{2}}=\left( {{6}^{2}}+{{\left( -3 \right)}^{2}} \right)$
But to find the magnitude of the complex number always put square roots and then do the square of the real and imaginary part and then add them.
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