Question

how to make 1/(3+4i) into re*i theta

Answer 1

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)

Answer 2

u gave me the answer for square root.. but i hav been asked to find exponential.
in the form re*itheta. thanks anyway

Answer 3

What's re?

Answer 4

r its the modulus.. and e is exponential

Answer 5

Do you think that first we should put it into a + bi form?

Answer 6

ya..by making use of its conjugate

Answer 7

So then we would have:
\[\frac{3}{25}-\frac{4}{25}i\]

Answer 8

mert is right

Answer 9

ya got it...n afterewards..finding modulus n theta.. ok thankss to both of u

Answer 10

yw

Answer 11

r e^iQ = r(cos Q + i sin Q)