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Suppose m = 2 + 6i, and |m+n| = 3sqrt10, where n is a complex number.

 

a. What is the minimum value of the modulus of n?

 

b. Provide one example of the complex number, n.

 May 6, 2020
 #1
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Suppose m = 2 + 6i, and |m+n| = 3sqrt10, where n is a complex number.
a. What is the minimum value of the modulus of n?
b. Provide one example of the complex number, n.

 

My attempt:

\(\begin{array}{|rcll|} \hline m&=& 2+6i \\ n&=& x+yi \\ \hline m+n &=& (2+x) + (6+y)i \\ \mathbf{|m+n|^2} = (2+x)^2 + (6+y)^2 &=& (3\sqrt{10})^2 \\ (2+x)^2 + (6+y)^2 &=& 9\cdot 10 \\ (2+x)^2 + (6+y)^2 &=& 90 \\ \hline \mathbf{|n|^2} &=& x^2+y^2 \\ \hline \end{array}\)

 

We wish to minimize \( f ( x , y ) = x^2 + y^2\)
subject to the constraint
\((2+x)^2+(6+y)^2=90\)

For the method of Lagrange multipliers, the constraint is \(g(x,y) =(2+x)^2+(6+y)^2 - 90 \)

 

hence
\(\Lambda(x,y,\lambda) = f(x,y)+\lambda g(x,y) \\ \Lambda(x,y,\lambda)= x^2 + y^2+\lambda\left[(2+x)^2+(6+y)^2 - 90\right] \)

 

Now we can calculate the gradient:

\(\begin{array}{|rcll|} \hline \dfrac{\partial \Lambda}{\partial x} &=& 2x+2\lambda(2+x) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{\partial \Lambda}{\partial y} &=& 2y+2\lambda(6+y) \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{\partial \Lambda}{\partial \lambda} &=& (2+x)^2+(6+y)^2 - 90 \\ \hline \end{array} \)

 

and therefore:

\(\begin{array}{|rcll|} \hline 2x+2\lambda(2+x) &=& 0 \quad | \quad : 2 \\ x+\lambda(2+x) &=& 0 \\ x+2\lambda+\lambda x &=& 0 \\ x(1+\lambda) &=& -2\lambda \\ \mathbf{x} &=& \mathbf{-\dfrac{ 2\lambda}{1+\lambda} } \\ \hline \end{array} \begin{array}{|rcll|} \hline 2y+2\lambda(6+y) &=& 0 \quad | \quad : 2 \\ y+\lambda(6+y) &=& 0 \\ y+6\lambda+\lambda y &=& 0 \\ y(1+\lambda) &=& -6\lambda \\ \mathbf{y} &=& \mathbf{-\dfrac{ 6\lambda}{1+\lambda} } \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline (2+x)^2+(6+y)^2 - 90 &=& 0 \\ (2+x)^2+(6+y)^2 &=& 90 \\ \left( 2-\dfrac{ 2\lambda}{1+\lambda} \right)^2+\left( 6-\dfrac{ 6\lambda}{1+\lambda} \right)^2 &=& 90 \\ \dfrac{ (2+2\lambda -2\lambda)^2 + (6+6\lambda-6\lambda)^2}{(1+\lambda)^2} &=& 90 \\ \dfrac{2^2+6^2}{(1+\lambda)^2} &=& 90 \\ \dfrac{40}{(1+\lambda)^2} &=& 90 \\ \ldots \\ (1+\lambda)^2 &=& \dfrac{40}{90} \\ (1+\lambda)^2 &=& \dfrac{4}{9} \\ 1+\lambda &=& \dfrac{2}{3} \\ \lambda &=& \dfrac{2}{3}-1 \\ \mathbf{\lambda} &=& \mathbf{-\dfrac{1}{3}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{-2\lambda}{1+\lambda} } \quad | \quad \mathbf{\lambda=-\dfrac{1}{3}} \\\\ x &=& \dfrac{ -2\left(-\dfrac{1}{3}\right)}{1-\dfrac{1}{3}} \\\\ x &=& \dfrac{ \dfrac{2}{3}}{\dfrac{2}{3}} \\\\ \mathbf{x} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{\dfrac{-6\lambda}{1+\lambda} } \quad | \quad \mathbf{\lambda=-\dfrac{1}{3}} \\\\ y &=& \dfrac{ -6\left(-\dfrac{1}{3}\right)}{1-\dfrac{1}{3}} \\\\ y &=& \dfrac{ \dfrac{6}{3}}{\dfrac{2}{3}} \\\\ y &=& \dfrac{ 2}{\dfrac{2}{3}} \\\\ \mathbf{y} &=& \mathbf{3} \\ \hline \end{array}\)

 

a.
The minimum value of the modulus of n is
\(\begin{array}{|rcll|} \hline && \mathbf{\sqrt{x^2+y^2}} \\ &=& \sqrt{1^2+3^2} \\ &=& \mathbf{\sqrt{10}} \\ \hline \end{array}\)

 

b.
One example of the complex number n is \(\mathbf{ 1+3i }\)

 

laugh

 May 6, 2020

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