Write (-1+2i)^(4) in simplest a+bi form. Answer:

Recognize Complex Number: To solve (-1+2i)^(4), we first need to recognize that we are raising a complex number to the fourth power. We can start by squaring (-1+2i) to simplify the expression step by step. (-1+2i)^(2) = (-1+2i) * (-1+2i) = 1 - 2i - 2i + 4i^2 Since i^2 = -1, we can substitute this into our expression. = 1 - 4i - 4(-1) = 1 - 4i + 4 = 5 - 4iSquare Complex Number: Now we have the square of the original complex number, (5 - 4i). To find the fourth power, we need to square (5 - 4i) again. (5 - 4i)^(2) = (5 - 4i) * (5 - 4i) = 25 - 20i - 20i + 16i^2 Again, we substitute i^2 with -1. = 25 - 40i + 16(-1) = 25 - 40i - 16 = 9 - 40iFind Fourth Power: We have now found (-1+2i)^(4) in the form of 9 - 40i, which is already in the simplest a+bi form.

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Q. Write

(-1+2 i)^{4}

in simplest

a+b i

form.

\newline

Answer:

Recognize Complex Number: To solve $(-1+2i)^{4}$ , we first need to recognize that we are raising a complex number to the fourth power. We can start by squaring $(-1+2i)$ to simplify the expression step by step. $(-1+2i)^{2} = (-1+2i) \times (-1+2i)$ $= 1 - 2i - 2i + 4i^{2}$ Since $i^{2} = -1$ , we can substitute this into our expression. $= 1 - 4i - 4(-1)$ $= 1 - 4i + 4$ $= 5 - 4i$
Square Complex Number: Now we have the square of the original complex number, $(5 - 4i)$ . To find the fourth power, we need to square $(5 - 4i)$ again. $\newline$ $(5 - 4i)^{2} = (5 - 4i) \times (5 - 4i)$ $\newline$ $= 25 - 20i - 20i + 16i^2$ $\newline$ Again, we substitute $i^2$ with $-1$ . $\newline$ $= 25 - 40i + 16(-1)$ $\newline$ $= 25 - 40i - 16$ $\newline$ $= 9 - 40i$
Find Fourth Power: We have now found $(-1+2i)^{4}$ in the form of $9 - 40i$ , which is already in the simplest $a+bi$ form.