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# e^(i*2*pi) * (-1)^98 * (343)^(1/3)

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e^(i*2*pi) * (-1)^98 * (343)^(1/3)

Guest Jun 8, 2017

#1
+1845
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I will assume for this problem that i is for the imaginary number. If I should assume otherwise, tell me. I will, of course, simplify the given expression:

 $$e^{i*2*\pi}*(-1)^{98}*343^{\frac{1}{3}}$$ We will use an imaginary number rule here stating that $$e^{ia\pi}=(-1)^a, \text{so}\hspace{1mm}e^{i2\pi}=(-1)^2=1$$. Of course, something multiplied by 1 is itself, so we are left with the other part. $$(-1)^{98}*343^{\frac{1}{3}}$$ -1 raised to an even power is always one, so this is another part of the multiplication that we can eliminate. $$343^{\frac{1}{3}}$$ $$a^{\frac{1}{3}}=\sqrt[3]{a}$$,so let's apply this rule, too. $$\sqrt[3]{343}=7$$ 7 is your answer.
TheXSquaredFactor  Jun 9, 2017
Sort:

#1
+1845
+1

I will assume for this problem that i is for the imaginary number. If I should assume otherwise, tell me. I will, of course, simplify the given expression:

 $$e^{i*2*\pi}*(-1)^{98}*343^{\frac{1}{3}}$$ We will use an imaginary number rule here stating that $$e^{ia\pi}=(-1)^a, \text{so}\hspace{1mm}e^{i2\pi}=(-1)^2=1$$. Of course, something multiplied by 1 is itself, so we are left with the other part. $$(-1)^{98}*343^{\frac{1}{3}}$$ -1 raised to an even power is always one, so this is another part of the multiplication that we can eliminate. $$343^{\frac{1}{3}}$$ $$a^{\frac{1}{3}}=\sqrt[3]{a}$$,so let's apply this rule, too. $$\sqrt[3]{343}=7$$ 7 is your answer.
TheXSquaredFactor  Jun 9, 2017

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