Question
1+7i(2−i)2=
Answer: Option A
:
A
1+7i(2−i)2=(1+7i)(3+4i)(3−4i)(3+4i)=−25+25i25= -1+i
Let z=x+iy = -1+i
∴ rcosθ =-1 And rsinθ =1 ∴ θ=3π4 and r=√2
Thus 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Alter: ∣∣∣1+7i(2−i)2∣∣∣=∣∣1+7i3−4i∣∣=√2
And arg (1+7i(2−i)2)=tan−17−tan−1(−43)
=tan−17+tan−1(43)=3π4
∴ 1+7i(2−i)2=√2 [cos3π4+isin3π4]
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:
A
1+7i(2−i)2=(1+7i)(3+4i)(3−4i)(3+4i)=−25+25i25= -1+i
Let z=x+iy = -1+i
∴ rcosθ =-1 And rsinθ =1 ∴ θ=3π4 and r=√2
Thus 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Alter: ∣∣∣1+7i(2−i)2∣∣∣=∣∣1+7i3−4i∣∣=√2
And arg (1+7i(2−i)2)=tan−17−tan−1(−43)
=tan−17+tan−1(43)=3π4
∴ 1+7i(2−i)2=√2 [cos3π4+isin3π4]
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