https://media.glynlyon.com/o_alg02_2011/7/t30.gif
who can help me?

Question

https://media.glynlyon.com/o_alg02_2011/7/t30.gif 
who can help me?

anonymous · Answer

$$28/61-(3 i)/61$$

anonymous · Answer

$$\frac{ 3+2i }{ 6+5i } \times \frac{ 6-5i }{ 6-5i }$$

anonymous · Answer

what jake is doin

anonymous · Answer

multiplying the top and bottom by the complex conjugate will allow you to eliminate the imaginary numbers in the denominator.

anonymous · Answer

i have 27i+(Something i dont know)/61
my answer but its not right

anonymous · Answer

On the parentheses   i dont know what is

anonymous · Answer

Top:  (3+2i)(6-5i) = 18 +12i -15i -10i^2 = 18 +10 - 3i = 28 - 3i

Bottom:  (6-5i)(6 + 5i) = 36 -30i + 30i -25i^2 = 36 + 25 = 51

anonymous · Answer

oops, 61, sorry

anonymous · Answer

thanks

anonymous · Answer

Corrected:

Top: (3+2i)(6-5i) = 18 +12i -15i -10i^2 = 18 +10 - 3i = 28 - 3i 
Bottom: (6-5i)(6 + 5i) = 36 -30i + 30i -25i^2 = 36 + 25 = 61

So overall, it's 28/61 - 3i/61

The whole idea when you have something like (a + bi) in the denominator is to multiply the top and bottom by (a - bi), leaving you a^2 + b^2 on the bottom, and putting the only imaginary numbers in the numerator.

anonymous · Answer

or the very first post.

anonymous · Answer

yeahh thanks  both