anonymous
  • anonymous
how to make 1/(3+4i) into re*i theta
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I'll do the first one for you and leat the second for you to apply what was done in the first one!! sqrt(3 - 4i) = a+bi 3 - 4i = (a+bi)^2 3 - 4i = (a^2 + 2abi -b^2) so 3 = a^2-b^2 and -4i = 2abi so -4i = 2abi a = 2/b then if 3 = a^2-b^2 subst. a = 2/b 3 = 4/b^2 - b^2 solve for b b^2 - 4/b^2 + 3 = 0 b^4 + 3b^2 - 4 = 0 now factor (b^2+4)(b^2 -1) = 0 roots are 2i, -2i, 1, -1 can't use 2i or -2i so b = either -1 or 1 now a = 2/b so a = 2/-1 = -2 or a = 2/1 = 2 for form a+ bi answers could be either (2 - 1i) or (-2+1i)
anonymous
  • anonymous
u gave me the answer for square root.. but i hav been asked to find exponential. in the form re*itheta. thanks anyway
Mertsj
  • Mertsj
What's re?

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anonymous
  • anonymous
r its the modulus.. and e is exponential
Mertsj
  • Mertsj
Do you think that first we should put it into a + bi form?
anonymous
  • anonymous
ya..by making use of its conjugate
Mertsj
  • Mertsj
So then we would have: \[\frac{3}{25}-\frac{4}{25}i\]
anonymous
  • anonymous
mert is right
anonymous
  • anonymous
ya got it...n afterewards..finding modulus n theta.. ok thankss to both of u
Mertsj
  • Mertsj
yw
experimentX
  • experimentX
r e^iQ = r(cos Q + i sin Q)

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