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sleet_krkn [62]
3 years ago
10

Let be the imag

inary unit.
What complex number in standard form does the following simplify to:

i^{425}+i^{14}+i^{-14}+i^{44}

Basically you are being asked to find A and B such that the following equality holds for real values A and B:
A+Bi=i^{425}+i^{14}+i^{-14}+i^{44}

You must show all work.
Have fun.
Mathematics
2 answers:
VLD [36.1K]3 years ago
8 0

Hey~freckledspots!\\----------------------

We~will~solve~for~i^{425}!

Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\

Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------

Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------

Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------

Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------

Answer:\\\large\boxed{-1+i}\\----------------------

Hope~This~Helped!~Good~Luck!

rodikova [14]3 years ago
3 0

Answer:

i - 1

Step-by-step explanation:

<u>Complex</u><u> </u><u>Number</u><u> </u><u>System</u><u> </u><u>Rules</u>

\sqrt{-1} = i \\ -1 = {i}^{2} \\ -i = {i}^{3} \\ 1 = {i}^{4} [Every \: multiple \: of \: four]

The Divisibilty Rule of 4 states that the<em> </em><em>last</em><em> </em><em>two</em><em> </em><em>digits</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>number</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>multiple</em><em> </em><em>of</em><em> </em><em>four</em><em>,</em><em> </em>so you will have this:

{i}^{425} + {i}^{14} + {i}^{-14} + {i}^{44} \\  \\ i - 1 - 1 + 1 = i - 1

I am joyous to assist you anytime.

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