i+(−i)=i+(−1)i=i+(i2)i=i+ii =i(1+i)for ix+iy=isquaring both the sides,x2−y2+2xiy=(i)2=ix2−y2=0 & 2xiy=ix=y & xy=21∴x=21 & y=21 OR x=−21 & y=−21are the solutions.So i+−i=i(1+i)=(1+i)(21+2i)=21−21+22i=2iAns=2i

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Text solutionVerified

i + (- i) = i + (- 1) i

= i + (i^{2}) i = i + i i

= i (1 + i)

for

i

x + i y = i

squaring both the sides,

x^{2} - y^{2} + 2 x i y = (i)^{2} = i

x^{2} - y^{2} = 0

2 x i y = i

x = y

x y = \frac{1}{2}

∴ x = \frac{1}{2}

y = \frac{1}{2}

x = - \frac{1}{2}

y = - \frac{1}{2}

are the solutions.

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

= \frac{1}{2} - \frac{1}{2} + 2 \frac{i}{2} = 2 i

A n s = 2 i

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Question Text	Find $i + - i$
Answer Type	Text solution:1
Upvotes	150

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Text solutionVerified

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