imaginary #'s?

Question

can you help please? i have 4 questions involving imaginary numbers.. can you solve them & tell me how you did it? thanks <3

1) simplify (1-3i)(-2+4i)


2) find the inverse of 2-5i

3) i^47 =

4) root -49 times root -4

number 4 please!!!!!!!!!!!!!!

Answer 1

(1)
(1 - 3i)(-2 + 4i)
= -2 + 4i + 6i -12i² . . by multiplying out the brackets
= -2 +4i + 6i + 12 . . . because i² = -1
= 10 + 10i . . . .gathering terms


(2)
inverse means 1/(2 - 5i)

Here we multiply the top and bottom of the fraction by the so-called complex conjugate of 2 - 5i i.e by 2 + 5i

so 1/(2 - 5i)
= 1*(2 + 5i) / {(2 - 5i)(2 + 5i)}
= (2 + 5i) / {4 + 25}
=(2/29) +(5/29)i


(3)
i² = -1, so i ^ 47= (i²)^23 = (-1)^23 = -1

So i ^ 47 = - i . . . . . it is just i^46 multiplied by an extra i

(4)
√-49 = √-1√49 = i*7 = 7i
√-4 = √-1√4 = i2 = 2i

So √-49 * √-4 = 7i * 2i = 14i² = -14

Answer 2

(1-3i)/(-2+4i) = (1-3i)(-2-4i)/20 (multiply by the conjugate)
= (-7+i)/10
2)1/2-5i= (2+5i)/29
3)i^1=i,i^2=-1, i^3=-i and i^4=1 and the cyclcle begins again
i^47=i^44*i^3 = i^3=-i
4) Work in polar form
-49 = 49 sqrt(-4) =2 so the product is 14 the product is 14 and -14.
In another ay
sqr-t49 = +-7i an sqrt(-4) =+-2i.Combining
Product =+-14
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