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kozerog [31]
3 years ago
12

Match the products of complex numbers with their standard forms.

Mathematics
2 answers:
erastova [34]3 years ago
8 0

<em>( 5 - 2i ) ( 4 - i )       </em><em>18 - 13i</em>

<em>( 2 - 5i ) ( 1 - 4i )     </em><em>-18 - 13i</em>

<em>( 1 - 4i ) ( 5 - 2i )      </em><em>-3 - 22i</em>

<em>( 4 + i ) ( 2 - 5i )     </em><em> 13 - 18i</em>

Alex17521 [72]3 years ago
3 0

Answer:

1) (5-2i)(4-i)=18-13i

2) (2-5i)(1-4i)=-18-13i

3) (1-4i)(5-2i)=-3-22i

4) (4+i)(2-5i)=13-18i    

Step-by-step explanation:

Given : Complex numbers

1) (5-2i)(4-i)

2) (2-5i)(1-4i)

3) (1-4i)(5-2i)

4) (4+i)(2-5i)

To find : The products of complex numbers with their standard forms.  

Solution :

We simply apply multiplicative property of brackets,

i.e, (a+b)(c+d)=ac+bc+bc+bd

Note - i^2=(\sqrt{-1})^2=-1

1) (5-2i)(4-i)

(5-2i)(4-i)=((5)(4)+(-2i)(4)+(-5)(i)+(-2i)(-i))

(5-2i)(4-i)=(20-8i-5i+2i^2)

(5-2i)(4-i)=(20-13i-2)

(5-2i)(4-i)=18-13i

2) (2-5i)(1-4i)

(2-5i)(1-4i)=((2)(1)+(-5i)(1)+(2)(-4i)+(-5i)(-4i))

(2-5i)(1-4i)=(2-5i-8i+20i^2)

(2-5i)(1-4i)=(2-13i-20)

(2-5i)(1-4i)=-18-13i

3) (1-4i)(5-2i)

(1-4i)(5-2i)=((5)(1)+(-4i)(5)+(-2i)(1)+(-2i)(-4i))

(1-4i)(5-2i)=(5-20i-2i+8i^2)

(1-4i)(5-2i)=(5-22i-8)

(1-4i)(5-2i)=-3-22i

4) (4+i)(2-5i)  

(4+i)(2-5i)=((4)(2)+(i)(2)+(4)(-5i)+(i)(-5i))

(4+i)(2-5i)=(8+2i-20i-5i^2)

(4+i)(2-5i)=(8-18i+5)

(4+i)(2-5i)=13-18i                

Therefore, following above are the required results.

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Hey there!

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What is the equation of 3p – 7 + p = 13
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How do I find the center &amp; radius, then write a standard equation for a,b, and c?
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Firstly , we will locate centres and radius in the figures

(a)

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now, we can use standard equation of circle

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we can plug values

we get

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(b)

We can see that

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now, we can use standard equation of circle

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we can plug values

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