1 + 7 i ( 2 − i ) 2 = Options: A . &nbsp √ 2 [ c o s 3 π 4 + i s i n 3 π 4 ] B . &nbsp √ 2 [ c o s π 4 + i s i n π 4 ] C . &nbsp [ c o s 3 π 4 + i s i n 3 π 4 ] D . &nbsp None of these

Answer: Option A : A 1 + 7 i ( 2 − i ) 2 = ( 1 + 7 i ) ( 3 + 4 i ) ( 3 − 4 i ) ( 3 + 4 i ) = − 25 + 25 i 25 = -1+i Let z=x+iy = -1+i ∴ rcos θ =-1 And rsin θ =1 ∴ θ = 3 π 4 and r= √ 2 Thus 1 + 7 i ( 2 − i ) 2 = √ 2 [ c o s 3 π 4 + i s i n 3 π 4 ] Alter: ∣ ∣ ∣ 1 + 7 i ( 2 − i ) 2 ∣ ∣ ∣ = ∣ ∣ 1 + 7 i 3 − 4 i ∣ ∣ = √ 2 And arg ( 1 + 7 i ( 2 − i ) 2 ) = tan − 1 7 − tan − 1 ( − 4 3 ) = tan − 1 7 + tan − 1 ( 4 3 ) = 3 π 4 ∴ 1 + 7 i ( 2 − i ) 2 = √ 2 [ c o s 3 π 4 + i s i n 3 π 4 ]

1+7i(2−i)2=

Discussion Forum : Complex Numbers

Question -

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

Options:

A .

\sqrt{2}

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

B .

\sqrt{2}

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

C .

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

D . None of these

Answer: Option A
:
A

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

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

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

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