Find square root of complex number -3-4iota
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I hope this answer help you
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KarupsK:
my solution is in simplest form.
but this method can not apply in all such type questions
Aahmmm I was exactly thinking the same .... but anywazzz thanks a lot
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let√-3-4i=x+yi
squaring both sides
-3-4i=(x+yi)²
-3-4i=x²+i²y³+2xyi
-3-4i=x²-y²+2xyi
so
x²-y²=-3
2xy=-4
(x²+y²)²=(x²-y²)²+(2xy)²
(x²+y²)²=(-3)²+(-4)²=9+16=25
x²+y²=5
x²-y²=-3
------------
2x²=2
x²=1 or x=±1
if x=1
2xy=-4
2(1)y=-4
y=-2
if x=-1
2(-1)y=-4
y=2
so
√-3-4i=±1-/+2
PLEASE SELECT MY ANSWERS AS BRAINLIEST
squaring both sides
-3-4i=(x+yi)²
-3-4i=x²+i²y³+2xyi
-3-4i=x²-y²+2xyi
so
x²-y²=-3
2xy=-4
(x²+y²)²=(x²-y²)²+(2xy)²
(x²+y²)²=(-3)²+(-4)²=9+16=25
x²+y²=5
x²-y²=-3
------------
2x²=2
x²=1 or x=±1
if x=1
2xy=-4
2(1)y=-4
y=-2
if x=-1
2(-1)y=-4
y=2
so
√-3-4i=±1-/+2
PLEASE SELECT MY ANSWERS AS BRAINLIEST
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