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2 years ago

Which is the additive inverse of the complex number -8+3i?

1 answer:

3 0

The additive inverse of a complex z is a complex number $-z,$ so that

$\mathsf{z+(-z)=-z+z=0+0i=0}$

Finding $-z:$

$\mathsf{z=-8+3i}\\\\\\ \mathsf{z+(-z)=0+0i}\\\\\ \mathsf{-8+3i+(-z)=0+0i}\\\\ \mathsf{-z=0+0i+8-3i}\\\\ \mathsf{-z=(0+8)+(0i-3i)}\\\\ \boxed{\begin{array}{c}\mathsf{-z=8-3i} \end{array}}\qquad\checkmark$

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Tags: <em>complex number additive inverse opposite algebra</em>

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Answer:

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Step-by-step explanation:

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2 years ago

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hodyreva [135]

Answer:

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Step-by-step explanation:

* Lets explain how to solve the problem

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2 years ago

Read 2 more answers

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gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$

Thus maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

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