Q: What is the polynomial roots to 1 5i and -3i?

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The four roots are:1 + 2i, 1 - 2i, 3i and -3i.

(3x - 1)(x - 4) so roots are 1/3 and 4

i is the Imaginary Unit, equal to sqrt(-1). So i and any real number multiplied by i will all be imaginary numbers. Here are some: i, -i, 5i, -3i, i*pi, etc.

1/(1+ 3i)

The answer to the question, as stated, is that the other root could be anything. However, if all the coefficients of the quadratic equation are real numbers, then the other root is 1 minus 3i.

Related questions

When adding and subtracting complex numbers, you can treat the "i" as any variable. For example, 5i + 3i = 8i, 5i -3i = 2i, etc.; (2 + 5i) - (3 - 3i) = (2 - 3) + (5 + 3)i = -1 + 8i.

There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.

2

-4-3i

The four roots are:1 + 2i, 1 - 2i, 3i and -3i.

1

It can have 1, 2 or 3 unique roots.

1/(2 + 5i) (multiply both the numerator and the denominator by 2 - 5i)= 1(2 - 5i)/(2 + 5i)(2 - 5i)= (2 - 5i)/(4 - 25i2) (substitute -1 for i2)= (2 - 5i)/(4 + 25)= (2 - 5i)/29= 2/29 - (5/29)i

1/(3+5i)=(3-5i)/((3+5i)(3-5i))=(3-5i)/(9+25)=(3-5i)/34

(3x - 1)(x - 4) so roots are 1/3 and 4

i is the Imaginary Unit, equal to sqrt(-1). So i and any real number multiplied by i will all be imaginary numbers. Here are some: i, -i, 5i, -3i, i*pi, etc.

1