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3 years ago

Divide (3+2i)/(-4+5i). express answer in standard form

1 answer:

8 0

$\dfrac{3+2i}{-4+5i}=\dfrac{(3+2i)(-4-5i)}{(-4+5i)(-4-5i)}=\dfrac{-12-15i-8i-10i^2}{(-4)^2-(5i)^2}\\\\=\dfrac{-12-23i+10}{16+25}=\dfrac{-2-23i}{41}=-\dfrac{2}{41}-\dfrac{23}{41}i$

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Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that

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The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)= $f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........$

Where the terms in f prime $f^{1}$ (a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

$f^{1}$ (a)= $f^{1}$ (-4)=-8/( $-4)^{2}$

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+ $f^{1}(-4)(x+4)/1!+ f^{11}(-4)(x+4)^{2}/2!.............$

$=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+ 24/(-4)^{3} (-4)(x+4)^{2}/2!.............$

=-2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$ /2!+...........

Learn more about taylor series at brainly.com/question/23334489

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