# Help with imaginary numbers?

I need help with imaginary numbers like with "i". I know i=sqr-1...but how do I do problems like:

1) 1/2+5i ?

I'm supposed to simplify it. How do I find out what 5i is?

2) 3+ isqr2/ 7-isqr2 ?

Relevance
• 1 decade ago

key point: the product of a complex number a + bi and its conjugate a - bi is the real number (a^2 + b^2)-- this technique is analogous to rationalizing a denominator--it makes denominators real

1) is this 1/(2 + 5i)?

if so, multiply the top and the bottom by the complex conjugate (2 - 5i) to give you: (2 - 5i) / 29, or 2/29 - 5i/29

If it's (1/2) + 5i, then this is already standard form... you could rearrange it to (1 + 10i)/2, but that isn't really simplifying

simplifying complex numbers usually means putting it into standard form a + bi

2) (3 + isqrt(2)) / (7 - isqrt(2))

multiply top and bottom by the conjugate of the denominator: (7 + isqrt(2))

(7 - isqrt(2))(7 + isqrt(2)) = 49 - i^2 (sqrt(2))^2 = 49 + 2 = 51

(3 + isqrt(2))(7 + isqrt(2)) / (51) =

[21 + 3isqrt(2) + 7isqrt(2) +i^2 (sqrt(2))^2 ] / 51=

(21 + 10 i sqrt(2) - 2 ) / 51 =

[19 + 10 i sqrt(2) ] / 51 =

19/51 + i10sqrt(2)/51