Find the value of $2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25}$ .

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Solution

$2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25}$
$= 2 (- 1) + 6 i (- 1) + 3 (i^{2})^{8} - 6 i (i^{2})^{9} + 4 i (i^{2})^{12}$
$= - 2 - 6 i + 3 (- 1)^{8} - 6 i (- 1)^{9} + 4 i (- 1)^{12}$
$= - 2 - 6 i + 3 + 6 i + 4 i$
$= 4 i + 1$

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Similar questions

Q. If

2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25} = x + i y

, then

(x, y) =

\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)} =

Q. Find the conjugate of

\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}

{⎛ ⎜ ⎝ \sqrt[3]{3} + ⎛ ⎜ ⎝ 3^{\frac{5}{6}} ⎞ ⎟ ⎠ i ⎞ ⎟ ⎠}^{3}

is an integer where

i = \sqrt{- 1}

. The value of the integer is equal to.

Find the conjuagates of the following complex numbers :

$(i) 4 - 5 i (i i) \frac{1}{3 + 5 i} (i i i) \frac{1}{1 + i} (i v) \frac{(3 - i)^{2}}{2 + i} (v) \frac{(1 + i) (2 + i)}{3 + i} (v i) \frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$

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Find the value of 2i2+6i3+3i16−6i19+4i25.

2i2+6i3+3i16−6i19+4i25=2(−1)+6i(−1)+3(i2)8−6i(i2)9+4i(i2)12=−2−6i+3(−1)8−6i(−1)9+4i(−1)12=−2−6i+3+6i+4i=4i+1

Find the value of $2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25}$ .

$2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25}$
$= 2 (- 1) + 6 i (- 1) + 3 (i^{2})^{8} - 6 i (i^{2})^{9} + 4 i (i^{2})^{12}$
$= - 2 - 6 i + 3 (- 1)^{8} - 6 i (- 1)^{9} + 4 i (- 1)^{12}$
$= - 2 - 6 i + 3 + 6 i + 4 i$
$= 4 i + 1$