How do you perform the operation and write the result in standard form given ${(4 + 5 i)}^{2}$ $(4+5i)^2$ ?

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Answer 1 · 2016-08-22T02:12:11

${(4 + 5 i)}^{2} = - 9 + 40 i$ $(4+5i)^2=-9+40i$

Whenever multiplication or division operations of complex numbers one need to always remember that $i^{2} = - 1$ $i^2=-1$ .

As such ${(4 + 5 i)}^{2}$ $(4+5i)^2$

= $4^{2} + 2 \times 4 \times 5 i + {(5 i)}^{2}$ $4^2+2×4×5i+(5i)^2$

= $16 + 40 i + 25 \times (- 1)$ $16+40i+25×(-1)$

= $16 + 40 i - 25$ $16+40i-25$

= $- 9 + 40 i$ $-9+40i$

How do you perform the operation and write the result in standard form given (4+5i)2(4+5i)^2?