Given: z= ( 1+2i 2+i ) ^ 2 = 1+4 i ^ 2 +4i 4+ i ^ 2 +4i = 3+4i 3+4i Now multiply (3 4i) to the numerator as well as denominator ∴ ( 3+4i) 3 ...

177

You visited us 177 times! Enjoying our articles? Unlock Full Access!

Algebra of Complex Numbers

Solve: 1+2i2...

Question

Solve:
${(\frac{1 + 2 i}{2 + i})}^{2}$

Open in App

Solution

Given: $z = {(\frac{1 + 2 i}{2 + i})}^{2} = \frac{1 + 4 i^{2} + 4 i}{4 + i^{2} + 4 i} = \frac{- 3 + 4 i}{3 + 4 i}$
Now multiply $(3 - 4 i)$ to the numerator as well as denominator
$∴ \frac{(- 3 + 4 i)}{3 + 4 i} \times \frac{3 - 4 i}{3 - 4 i} = \frac{- 9 + 12 i + 12 i - 16 i^{2}}{9 - 16 i^{2}} = \frac{- 9 + 24 i + 16}{9 + 16} = \frac{7 + 24 i}{25} ∴ z = \frac{7}{25} + i \frac{24}{25}$

Suggest Corrections

Similar questions

{(\frac{2 i}{1 + i})}^{2}

\frac{1 + 2 i}{1 - {(1 - i)}^{2}}

{(\frac{2 i}{1 + i})}^{2}

\frac{(2 - 3 i)}{(2 - 2 i)}

i^{2} = - 1

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

Join BYJU'S Learning Program