Question

prove...................................​

Answer 1

Answer:

$\bf{(i^{17}-(\frac{1}{i})^{34})^2=2i}$

Step-by-step explanation:

$\text{Prove that: (i^{17}-(\frac{1}{i})^{34})^2=2i }$

Concept used:

$\boxed{i^2=-1\:\:\:\implies\:\frac{1}{i}=-i}$

Now,

$(i^{17}-(\frac{1}{i})^{34})^2$

$=(i^{16}.i^1-(-i)^{34})^2$

using $\boxed{i^{4n}=1,n\:is\:an\:integer}$

$=(1.i^1-(-i)^{34})^2$

$=(i-i^{34})^2$

$=(i-i^{32}.i^2)^2$

$=(i-1.(-1))^2$

$=(i+1)^2$

using $\boxed{(a+b)^2=a^2+b^2+2ab}$

$=i^2+1^2+2(i)(1)$

$=-1+1+2(i)(1)$

$=2i$

$\implies\:\bf{(i^{17}-(\frac{1}{i})^{34})^2=2i}$

Answer 2

Answer:

Step-by-step explanation:mark me i needed

The answer is here,

Given that,

= > \: \frac{a + ib}{c + id} = p + iq

We can replace the " i " as " -i ".

= > \frac{a - ib}{c - id} = p - iq