What is ((2i^5+3i^{17})/(1+i))^2 ?

The solution to ((2i^5+3i^{17})/(1+i))^2 is 25/2 i

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simplify ((2i^5+3i^{17})/(1+i))^2

simplify

(\frac{2 i ^{5} + 3 i ^{17}}{1 + i})^{2}

\frac{25}{2} i

Solution steps

(\frac{2 i ^{5} + 3 i ^{17}}{1 + i})^{2}

Simplify

\frac{2 i ^{5} + 3 i ^{17}}{1 + i} : \frac{5 i ( 1 - i )}{2}

= (\frac{5 i ( 1 - i )}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 5 i ( 1 - i ) ) ^{2}}{2 ^{2}}

Simplify

(5 i (1 - i))^{2} : 50 i

= \frac{50 i}{2 ^{2}}

2^{2} = 4

= \frac{50 i}{4}

Cancel the common factor:

2

= \frac{25 i}{2}

Rewrite in standard complex form:

\frac{25}{2} i

= \frac{25}{2} i

lo g_{x} (\frac{( x + 4 ) ^{2}}{2 x + 5}) < 0

q^{2} + 7 q + 10

9 y + 5 y + 11 = 16 - 6 y

5 t^{4} + 5 t^{3} - 210 t^{2}

\frac{2 x - 3}{5} - \frac{x}{2} = 1 - \frac{x - 1}{4}

What is ((2i^5+3i^{17})/(1+i))^2 ?
The solution to ((2i^5+3i^{17})/(1+i))^2 is 25/2 i