1+7i/(2 i)^2=(1+7i)(3+4i)/(3 4i)(3+4i)= 25+25i/25= 1+i Let z=x+iy = 1+i ∴ rcosθ = 1 And rsinθ =1 ∴ θ=3π/4 and r=√(2) Thus 1+7i/(2 i)^2=√(2) [cos3π/4 +isin3 ...

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Algebra of Complex Numbers

1+7i/2-i2=

Question

$\frac{1 + 7 i}{(2 - i)^{2}}$ =

$\sqrt{2}$ $[c o s \frac{3 π}{4} + i s i n \frac{3 π}{4}]$

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$\sqrt{2}$ $[c o s \frac{π}{4} + i s i n \frac{π}{4}]$

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$[c o s \frac{3 π}{4} + i s i n \frac{3 π}{4}]$

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None of these

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Solution

The correct option is A
$\sqrt{2}$ $[c o s \frac{3 π}{4} + i s i n \frac{3 π}{4}]$

$\frac{1 + 7 i}{(2 - i)^{2}}$ = $\frac{(1 + 7 i) (3 + 4 i)}{(3 - 4 i) (3 + 4 i)}$ = $\frac{- 25 + 25 i}{25}$ = -1+i

Let z=x+iy = -1+i

$∴$ rcos $θ$ =-1 And rsin $θ$ =1 $∴$ $θ$ = $\frac{3 π}{4}$ and r= $\sqrt{2}$

Thus $\frac{1 + 7 i}{(2 - i)^{2}}$ = $\sqrt{2}$ $[c o s \frac{3 π}{4} + i s i n \frac{3 π}{4}]$

Alter: $∣ ∣ ∣ \frac{1 + 7 i}{(2 - i)^{2}} ∣ ∣ ∣$ = $∣ ∣ \frac{1 + 7 i}{3 - 4 i} ∣ ∣$ = $\sqrt{2}$

And arg $(\frac{1 + 7 i}{(2 - i)^{2}})$ = ${tan}^{- 1} 7 - {tan}^{- 1} (\frac{- 4}{3})$

= ${tan}^{- 1} 7 + {tan}^{- 1} (\frac{4}{3})$ = $\frac{3 π}{4}$

$∴$ $\frac{1 + 7 i}{(2 - i)^{2}}$ = $\sqrt{2}$ $[c o s \frac{3 π}{4} + i s i n \frac{3 π}{4}]$

Suggest Corrections

1+7i(2−i)2=

$\frac{1 + 7 i}{(2 - i)^{2}}$ =