Question

how do you solve this z^2=-12+16i for z in a +bi form ?

Answer 1

-12+16i is in the form a+bi

a = -12
b = 16

convert this to r*( cos(t) + i*sin(t) ) form

where r = radius, t = angle theta

r = sqrt( a^2 + b^2 )

r = sqrt( (-12)^2 + (16)^2 )

r = 20

t = arctan(b/a)

t = arctan(16/(-12))

t = -53.1301

Answer 2

so

-12+16i

converts to

20*( cos(-53.1301) + i*sin(-53.1301) )

and that's the same as

20*cis(-53.1301)

Answer 3

thanks jim, you're the greatest B)

Answer 4

using euler's formula, we can say

20*cis(-53.1301) = 20*e^(-53.1301i)

so we now know this

z^2=-12+16i

is the same as

z^2=20*e^(-53.1301i)

now take the square root of both sides

z = +-sqrt(20*e^(-53.1301i))

z = +-sqrt(20)*sqrt(e^(-53.1301i))

z = +-2*sqrt(5)*e^(-53.1301i/2)

z = +-2*sqrt(5)*e^(-26.56505)

then convert back to r*( cos(t) + i*sin(t) ) form

Answer 5

z = +-2*sqrt(5)*e^(-26.56505)

z = +-2*sqrt(5)*cis(-26.56505)

z = +-2*sqrt(5)*[ cos(-26.56505) + i*sin(-26.56505) ]

now convert to a+bi form

z = +-2*sqrt(5)*[ cos(-26.56505) + i*sin(-26.56505) ]

z = +-2*sqrt(5)*[ 0.894427 - 0.447214i]

z = 2*sqrt(5)*[ 0.894427 - 0.447214i] or z = -2*sqrt(5)*[ 0.894427 - 0.447214i]

z = 2*sqrt(5)*0.894427 - 2*sqrt(5)*0.447214i or z = -2*sqrt(5)* 0.894427 + 2*sqrt(5)*0.447214i


z = 1.788854*sqrt(5) - 0.894428*sqrt(5)*i or z = -1.788854*sqrt(5) + 0.894428*sqrt(5)*i

Answer 6

hold on, must have done something wrong, one sec

Answer 7

i thin kyou just got it backwards! but thanks jim!

Answer 8

no i found a better way, one sec

Answer 9

alright :)

Answer 10

Let z = a+bi


z^2 = (a+bi)^2


z^2 = (a+bi)(a+bi)


z^2 = a(a+bi)+bi(a+bi)


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


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


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


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


Real Part: a^2 - b^2


Imaginary Part: 2ab


Real part in -12+16i is -12


so a^2 - b^2 = -12


Imaginary part is 16, so 2ab = 16


2ab = 16


ab = 8


b = 8/a


------------------


a^2 - b^2 = -12


a^2 - (8/a)^2 = -12


a^2 - 64/(a^2) = -12


q - 64/q = -12 ... let q = a^2


q^2 - 64 = 12q


q^2 + 12q - 64 = 0


(q-4)(q+16) = 0


q-4 = 0 or q+16 = 0


q = 4 or q = -16


a^2 = 4 or a^2 = -16


a = 2, a = -2, a = 4i, a = -4i



If a = 2, then


b = 8/a


b = 8/2


b = 4


So one solution is z = 2+4i


If a = -2, then


b = 8/a


b = 8/(-2)


b = -4


So another solution is z = -2-4i


If you are restricting a and b to be real numbers, then those are the only two solutions

Answer 11

So again, the only two solutions are 2+4i and -2-4i if you are restricting a and b to be real numbers

Answer 12

yeah i did get the first part backwards, idk how, but i fixed it using this better approach