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Morgarella [4.7K]
3 years ago
7

In an Argand diagram, the point P represents the complex number z, where z = x + iy. Given that z + 2 = λi(z + 8), where λ is a

real parameter, find the Cartesian equation of the locus of P as λ varies. If also z = μ(4 + 3i), where λ is real, prove that there is only one possible position for P
Business
1 answer:
Scilla [17]3 years ago
8 0

Answer:

The answer is   "\boxed{z= \frac{-16}{5}- \frac{-12}{15}i}".

Explanation:

As the problem stands  

At the point of P, it is the complex number z in the Diagram of Argand and z = X+iy.  

We have said this: (z+2)= \lambda i (z +8) .... (i)

where the \lambda parameter is a true  

The conceptual equation of the locus P varies between z = x+iy  \ \ \ to \ \ \ \lambda

And in equation mentioned above.

x+iy+2=\lambda i(x+iy+8) \\\\x+iy+2= \lambda xi+ \lambda i^2y+\lambda 8i\\\\ x+2+iy=-y \lambda +i(x+8)\lambda\\\ compare \ real \ and \ imaginary\  part \\\\\ x+2 = -y\lambda \\\\y= (x+8) \lambda\\\\   \lambda  = \frac{x+2}{-y}  \\ \\ \lambda = \frac{y}{x+8}

y^2= -x^2-10x-16 ....(ii)\\\\z= \mu (4+3i)....(iii)\\\\\ z= x+iy \\\\x+iy = 4\mu + 3 \mu i \\\\x= 4\mu \\\\y= 3\mu

put the value of x, y  in equation (ii) we get:

5\mu +4=0\\\\\mu = \frac{-4}{5} \\\\

to put the of \mu in equation (iii) we get:

z= \frac{-4}{5} (4+3i) \\\\ \boxed{z= \frac{-16}{5}- \frac{-12}{15}i} \\

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Discount on issue                     20000                                      

Annual discount                        4000

amortization

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Interest expense                         64000

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Interest payable                           60,000

Cash                                                                                60,000

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