Question

Express (3+7i)^2 in the form of a+ib​

Answer 1

${(3 + 7i)}^{2} \\ \\ = ( {3)}^{2} + 2.3.7i + ( {7i)}^{2} \\ \\ = 9 + 42i + 49 {i}^{2} \\ \\ = 9 + 42i + 49 \times - 1 \\ \\ = 9 + 42i - 49 \\ \\ = 9 - 49 + 42i \\ \\ = - 40 + 42i$

I hope this helps.

Answer 2

To express:

$(3+7i)^{2}$ in the form of $a+ib$

Step-by-step explanation:

Now $(3+7i)^{2}$

• Use algebraic identity $(a+b)^{2}=a^{2}+2ab+b^{2}$ to expand the expression.

$=(3)^{2}+2\times 3\times 7i+(7i)^{2}$

$=9+42i+49i^{2}$

$=9+42i+49(-1)$, since $i^{2}=-1$

$=9+42i-49$

$=9-49+42i$

$=-40+42i$

Final answer:

Thus $(3+7i)^{2}=-40+42i$ which is in the form of $a+ib$ where $a=-40,b=42$.

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