The magnitude of the right side is 4 so the magnitude of z must be 2. That is not true of any of the choices, so none of the choices are roots (possible values of z).
You can reach the same conclusion by squaring each choice and seeing what you get for the real and imaginary terms. None come close to 1-i*sqrt3. For example,
(-0.22+1.24i)^2 = (-0.22)^2 -(1.24)^2
-2*(0.22)(1.24)i = -1.489 -0.546 i
Which of the following is not a root of z^2= 1-square root of 3i to the nearest hundredth.
a. -0.22+1.24i
b. -0.97-0.81i
c. 1.02-0.65i
d. 1.18-0.43i
Iam not sure how to do this problem. Can someone help me? Thanks
2 answers
another way of doing this is to convert to polar
1-isqrt3=4@arctan-3/1
sqrt above=+- [email protected]
arctan -1.5=-56deg appx
so roots are
2@-56deg and 2@124deg or
2cos56-i2sin56 or 2cos124+isin124 or
1.18-1.65i or -1.18 +i1.65
which indicates none are roots.
1-isqrt3=4@arctan-3/1
sqrt above=+- [email protected]
arctan -1.5=-56deg appx
so roots are
2@-56deg and 2@124deg or
2cos56-i2sin56 or 2cos124+isin124 or
1.18-1.65i or -1.18 +i1.65
which indicates none are roots.
1 answer