Question

If the value of ${{i}^{2}}=-1$, then calculate the value of $3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}$.
(a) $-4-i$
(b) $-2-i$
(c) $2+i$
(d) $4+i$
(e) $6+2i$

Answer 1

To solve this question we will assume variables a, b, c to $3{{i}^{2}},{{i}^{3}}$ & $-{{i}^{4}}$ and we will use the identity that ${{i}^{2}}=-1$. Finally we will add them up to get the result.

Complete step-by-step solution:
We are given that the value of ${{i}^{2}}=-1$, then the value of ${{i}^{3}}$, ${{i}^{4}}$ can be calculated separately.
Firstly we will calculate the value of ${{i}^{3}}$ and ${{i}^{4}}$, thus proceed to calculate the value of $3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}$.
We have, ${{i}^{2}}=-1$.
Then, Let $a=3{{i}^{2}},b={{i}^{3}},c=-{{i}^{4}}$.
We have to calculate the value of a + b + c,
Because, ${{i}^{2}}=-1$.

& \Rightarrow 3{{i}^{2}}=\left( 3 \right)\left( -1 \right) \\\ & \Rightarrow 3{{i}^{2}}=-3 \\\ \end{aligned}$$ Therefore, $a = -3$ ------- (1) Now consider b. $$b={{i}^{3}}$$ We have, $${{i}^{2}}=-1$$. Multiplying ‘i’ both sides of the above equation, $$\Rightarrow {{i}^{3}}=-i$$ $$\Rightarrow b = -i $$ ---------- (2) Now compute, $$c=-{{i}^{4}}$$. We have, $${{i}^{2}}=-1$$. Multiplying, $${{i}^{2}}=-1$$ on both sides we have, $$\begin{aligned} & {{i}^{4}}=\left( -1 \right)\left( -1 \right) \\\ & \Rightarrow {{i}^{4}}=1 \\\ \end{aligned}$$ Now, $$c=-{{i}^{4}}=-1$$. Hence, $c = -1$ –------ (3) Now a + b + c, using (1), (2) & (3) we have, $$a+b+c=-3-i-1=-4-i$$, which is option (a). **Therefore, $$3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}=-4-i$$, option (a) is correct.** **Note:** Another way to solve this question can be directly. Substituting, $${{i}^{2}}=-1$$, $${{i}^{4}}=1$$ & $${{i}^{3}}=-i$$ to get the result, then answer would come as $$3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}=3\left( -1 \right)+\left( -i \right)-1=-4-i$$, option (a).