Question

(a+bi)^2=i what a and b satisfies the equation?

Answer 1

square it first

Answer 2

(a+bi)^2=i

(a+bi)(a+bi)=i

a^2+2abi +b^2i^2 = i

a^2+2abi +b^2(-1) = i

a^2+2abi - b^2 = i

(a^2-b^2) + 2ab*i = 0 + 1*i

So this means

a^2 - b^2 = 0

2ab = 1

Answer 3

what u can do is expand the left then simplify... then match up the real parts and the imaginary parts...

Answer 4

You now have a system of two equations with 2 unknowns. So you can solve for both 'a' and 'b' using this system.

Answer 5

could you please solve I do not understand fully

Answer 6

Solve the second equation for any variable (I'm choosing b)

2ab = 1

b = 1/(2a)

Now plug this into the other equation

a^2 - b^2 = 0

a^2 - (1/(2a))^2 = 0

a^2 - 1/(4a^2) = 0

4a^4 - 1 = 0

I'll let you finish

Answer 7

another way is express i in polar form

\[ i= e^{i \frac{\pi}{2}} \]
if you want sqrt(i):
\[ i^{\frac{1}{2}}= (e^{i \frac{\pi}{2}} )^\frac{1}{2}=e^{i \frac{\pi}{4}} \]

or
cos(pi/4) + i sin(pi/4)
sqrt(2)/2 + i sqrt(2)/2

Answer 8

phi could you explain your way?

Answer 9

do you know
e^ix = cos(x) + i*sin(x)
?
if x= pi/2, cos(pi/2)=0 and sin(pi/2)= 1
so
e^(i*pi/2) = i
to take the square root, raise e^(i*pi/2) to the 1/2 power (multiply exponents)
change back to "rectangular form" using e^ix = cos(x) + i*sin(x)

Answer 10

so a+bi is just another way of writing cos(x) + i*sin (x)?

Answer 11

almost. a+bi --> sqrt(a^2+b^2) * e^(atan(b/a))
See http://en.wikipedia.org/wiki/Complex_number#Polar_form

it is helpful to know the magnitude of i is 1.

Answer 12

Thanks so much!

Answer 13

If you don't know this, then stay with rectangular form and solve for a and b as jt showed.