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4 times Find the value of i^703 . 4 x & A & 1 4 x & B & 1 4 x & C & -1 4 x & D & -1

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4 times Find the value of i^703 . 4 x & A & 1 4 x & B & 1 4 x & C & -1 4 x & D & -1

4 times Find the value of i^703 . 4 x & A & 1 4 x & B & 1 4 x & C & -1 4 x & D & -1

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MeredithElite · Tutor for 8 years

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# Explanation<br /><br />To find the value of \( i^{703} \), we need to understand the properties of the imaginary unit \( i \). The imaginary unit \( i \) is defined as \( i = \sqrt{-1} \). The powers of \( i \) follow a cyclical pattern:<br /><br />\[<br />\begin{align*}<br />i^1 &= i, \\<br />i^2 &= -1, \\<br />i^3 &= -i, \\<br />i^4 &= 1, \\<br />i^5 &= i, \\<br />i^6 &= -1, \\<br />i^7 &= -i, \\<br />i^8 &= 1, \\<br />&\vdots<br />\end{align*}<br />\]<br /><br />From this, we see that the powers of \( i \) repeat every 4 terms. Therefore, to find \( i^{703} \), we can use modular arithmetic to determine the equivalent power within the first cycle of 4 terms.<br /><br />We calculate the remainder when 703 is divided by 4:<br /><br />\[<br />703 \mod 4<br />\]<br /><br />Performing the division:<br /><br />\[<br />703 \div 4 = 175 \text{ remainder } 3<br />\]<br /><br />Thus,<br /><br />\[<br />703 \equiv 3 \pmod{4}<br />\]<br /><br />This means that \( i^{703} \) is equivalent to \( i^3 \). From the cyclical pattern, we know:<br /><br />\[<br />i^3 = -i<br />\]<br /><br />Therefore, \( i^{703} = -i \).<br /><br /># Answer<br /><br />C
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